How do you use the Ratio Test on the series #sum_(n=1)^oo9^n/n# ?

Answer 1

Let

#a_n={9^n}/{n}#. #Rightarrow a_{n+1}={9^{n+1}}/{n+1}#

By Ratio Test,

#lim_{n to infty}|{a_{n+1}}/{a_n}|=lim_{n to infty}|{{9^{n+1}}/{n+1}}/{{9^n}/{n}}|#

by cancelling out common factors,

#=lim_{n to infty}|{9n}/{n+1}|#
by pulling 9 out of the limit and dividing the numerator and denominator by #n#,
#=9lim_{n to infty}|{1}/{1+1/n}|=9cdot 1/{1+0}=9 ge 1#

Hence, the series diverges.

I hope that this was helpful.

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Answer 2

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[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

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[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

AfterTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

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[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

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To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifyingTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying,To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_nTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, ifTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limitTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \fracTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit isTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less thanTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^nTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1,To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{nTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}\To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely.To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}).To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If itTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). IfTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it'sTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If theTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greaterTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limitTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit isTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is lessTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 orTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less thanTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or theTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefinedTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1,To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series divergesTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges.To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges.To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. IfTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. IfTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If itTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limitTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it'sTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equalsTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greaterTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater thanTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1,To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1, theTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than 1 orTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1, the testTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than 1 or the limitTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1, the test isTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than 1 or the limit is undefined,To use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1, the test is inconTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than 1 or the limit is undefined, theTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1, the test is inconclusiveTo use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than 1 or the limit is undefined, the seriesTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1, the test is inconclusive.To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than 1 or the limit is undefined, the series diverTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1, the test is inconclusive.To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than 1 or the limit is undefined, the series divergesTo use the Ratio Test on the series ( \sum_{n=1}^{\infty} \frac{9^n}{n} ), you would take the limit as ( n ) approaches infinity of the absolute value of the ratio of the ( (n+1) )th term to the ( n )th term:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{9^{n+1}}{n+1}}{\frac{9^n}{n}} \right| ]

After simplifying, if the limit is less than 1, the series converges absolutely. If it's greater than 1 or the limit is undefined, the series diverges. If the limit equals 1, the test is inconclusive.To use the Ratio Test on the series (\sum_{n=1}^{\infty} \frac{9^n}{n}), you would calculate the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

where (a_n = \frac{9^n}{n}). If the limit is less than 1, the series converges. If it's greater than 1 or the limit is undefined, the series diverges. If the limit is equal to 1, the test is inconclusive, and another test may be needed.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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