How do you apply the ratio test to determine if #Sigma (5^n)/(6^n5^n)# from #n=[1,oo)# is convergent to divergent?
The series:
is convergent.
Evaluate the ratio:
so:
which proves the series to be convergent.
By signing up, you agree to our Terms of Service and Privacy Policy
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.

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}} ]

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})} ]

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 ]

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

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

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.
By signing up, you agree to our Terms of Service and Privacy Policy
When evaluating a onesided 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 onesided 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 onesided 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 onesided 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.
 How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma (1)^(n+1)/sqrtn# from #[1,oo)#?
 How do you apply the ratio test to determine if #Sigma (1*3*5* * * (2n1))/(n!)^2# from #n=[1,oo)# is convergent to divergent?
 How do you show that the series #1+sqrt2+root3(3)+...+rootn(n)+...# diverges?
 What is the Harmonic Series?
 How do I determine if the alternating series #sum_(n=1)^oo(1)^n/sqrt(3n+1)# is convergent?
 98% accuracy study help
 Covers math, physics, chemistry, biology, and more
 Stepbystep, indepth guides
 Readily available 24/7