How do you apply the ratio test to determine if #Sigma (5^n)/(6^n-5^n)# from #n=[1,oo)# is convergent to divergent?

Answer 1

The series:

#sum_(n=1)^oo 5^n/(6^n-5^n)#

is convergent.

Evaluate the ratio:

#abs(a_(n+1)/a_n) = abs ( (5^(n+1)/(6^(n+1)-5^(n+1)))/(5^n/(6^n-5^n)))#
#abs(a_(n+1)/a_n) = (5^(n+1)/5^n ) ( (6^n-5^n) / (6^(n+1)-5^(n+1)))#
#abs(a_(n+1)/a_n) = 5*6^n( (1-(5/6)^n) / (6^(n+1)-5^(n+1)))#
#abs(a_(n+1)/a_n) = 5/6 *6^(n+1)( (1-(5/6)^n) / (6^(n+1)-5^(n+1)))#
#abs(a_(n+1)/a_n) = 5/6 ( (1-(5/6)^n) / ((6^(n+1)-5^(n+1) )/ 6^(n+1)))#
#abs(a_(n+1)/a_n) = 5/6 ( (1-(5/6)^n) / (1-(5/6)^(n+1) ))#
Now as #5/6 < 1#
#lim_(n->oo) (5/6)^n = lim_(n->oo) (5/6)^(n+1) = 0#

so:

#lim_(n->oo) abs(a_(n+1)/a_n) = 5/6 lim_(n->oo) ( (1-(5/6)^n) / (1-(5/6)^(n+1) )) = 5/6 < 1#

which proves the series to be convergent.

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Answer 2

To apply the ratio test to the series (\sum_{n=1}^{\infty} \frac{5^n}{6^n - 5^n}), you need to compute the limit of the absolute value of the ratio of consecutive terms as (n) approaches infinity.

  1. Compute the ratio of consecutive terms: [ \frac{a_{n+1}}{a_n} = \frac{\frac{5^{n+1}}{6^{n+1} - 5^{n+1}}}{\frac{5^n}{6^n - 5^n}} ]

  2. Simplify the expression: [ \frac{a_{n+1}}{a_n} = \frac{5^{n+1}(6^n - 5^n)}{5^n(6^{n+1} - 5^{n+1})} ]

  3. Take the limit as (n) approaches infinity: [ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{5^{n+1}(6^n - 5^n)}{5^n(6^{n+1} - 5^{n+1})} \right| ]

[ = \lim_{n \to \infty} \left| \frac{5(6^n - 5^n)}{6 \cdot 6^n - 5 \cdot 5^n} \right| ]

  1. Simplify further: [ = \lim_{n \to \infty} \left| \frac{5(1 - (\frac{5}{6})^n)}{6 - (\frac{5}{6})^n} \right| ]

  2. As (n) approaches infinity, the term ((\frac{5}{6})^n) approaches zero. Therefore, the limit becomes: [ = \frac{5}{6} ]

  3. Evaluate the limit. If the limit is less than 1, the series converges. If it's greater than 1 or undefined, the series diverges.

Since the limit (\frac{5}{6}) is less than 1, by the ratio test, the series (\sum_{n=1}^{\infty} \frac{5^n}{6^n - 5^n}) converges.

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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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