Use ratio test to find value of $\sum_{k=1}^\infty\frac{(\frac{1}{10})^{k+1}}{k(k+1)}$

Asked Dec 06 '19 at 17:06

Active Dec 06 '19 at 17:37

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I know the series $$\sum_{k=1}^\infty \frac{(\frac{1}{10})^{k+1}}{k(k+1)}$$ is equal to $$\frac{1}{10}-\frac{9}{10}\ln \Big(\frac{10}{9}\Big)$$ but how would I use the ratio test to show this is true?

I also know that: $$ (1-x)\ln(1-x)=\sum^\infty_{k=1}\frac{x^{k+1}}{k}-\sum^\infty_{k=1}\frac{x^k}{k} =\sum_{k=1}^\infty \frac{x^{k+1}}{k(k+1)}$$

edited Dec 06 '19 at 17:37

Kenta S

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asked Dec 06 '19 at 17:06

undergrad2020

The ratio test does not find the value of a series. It finds only whether the series converges or not. – GEdgar Dec 06 '19 at 17:36
I was told that it would work. What test should I use to show this converges to that value? – undergrad2020 Dec 06 '19 at 17:46
Use your last line, and find what to plug in for $x$ to get your series. But I would not call it a "test". – GEdgar Dec 06 '19 at 17:48
Okay, thank you. – undergrad2020 Dec 06 '19 at 17:49

Use ratio test to find value of $\sum_{k=1}^\infty\frac{(\frac{1}{10})^{k+1}}{k(k+1)}$

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