How do you use the Ratio Test on the series $sum_(n=1)^oo(-10)^n/(4^(2n+1)(n+1))$ ?

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Answer 1 · 2014-09-04T00:24:27

By Ratio Test, the posted series converges absolutely.

By Ratio Test:

$lim_{n to infty}|a_{n+1}/a_n|=lim_{n to infty}|{(-10)^{n+1}}/{4^{2n+3}(n+2)}cdot{4^{2n+1}(n+1)}/{(-10)^n}|$

By canceling out common factors:

$=lim_{n to infty}|{-10(n+1)}/{4^2(n+2)}|$

since $|{-10}/4^2|=5/8$ , we have:

$=5/8lim_{n to infty}(n+1)/(n+2)$

by dividing the numerator and the denominator by $n$ ,

$=5/8 lim_{n to infty}{1+1/n}/{1+2/n}=5/8cdot 1=5/8<1$

Hence, $sum_{n=1}^{infty}{(-10)^n}/{4^{2n+1}(n+1)}$ is absolutely convergent.

How do you use the Ratio Test on the series sum_(n=1)^oo(-10)^n/(4^(2n+1)(n+1))sum_(n=1)^oo(-10)^n/(4^(2n+1)(n+1)) ?