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I am trying to evaluate the following sum:

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)5^n}$$

So far I have written the sum as

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)5^n} = \sum_{n=1}^{\infty} \left ( \frac{1}{n} - \frac{1}{n+1} \right ) \frac{1}{5^n} = \sum_{n=1}^{\infty} \frac{1}{n5^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1)5^n}$$

I am stuck and I have not been able to find any similar examples. Wolfram Alpha gives the result

$$\sum_{n=1}^{\infty} \frac{1}{n(n+1)5^n} = 1-4 \log(5/4)$$

I feel as though I should be writing the sum as an integral and evaluating the integral, but I do not know how to proceed. Any help appreciated. Thanks.

clarence_atwater
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3 Answers3

10

Hint: After your good start use $\ \displaystyle \sum_{n=1}^\infty \frac {x^n}n=-\ln(1-x)$.

Raymond Manzoni
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5

$$\sum_{k=1}^{\infty} \dfrac{x^k}k = \sum_{k=1}^{\infty} \left( \int_0^x t^{k-1}dt \right)= \sum_{k=0}^{\infty} \left(\int_0^x t^{k}dt \right) = \int_0^x \left(\sum_{k=0}^{\infty}t^k\right) dt = \int_0^x \dfrac{dt}{1-t} = -\log(1-x)$$ Hence, $$\sum_{k=1}^{\infty} \dfrac1{k\cdot5^k} = - \log(4/5)$$ $$\sum_{k=1}^{\infty} \dfrac1{(k+1)\cdot5^k} = 5 \times \sum_{k=1}^{\infty} \dfrac1{(k+1)\cdot5^{k+1}} = 5 \times \left(- \log(4/5) - \dfrac15\right) = -1 - 5 \log(4/5)$$

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$$\begin{align} \sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=1}^\infty \frac{1}{(n+1)5^n} &= \sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=2}^\infty \frac{1}{n5^{n-1}} \\ &=\sum_{n=1}^\infty \frac{1}{n5^n} - \sum_{n=1}^\infty \frac{1}{n5^{n-1}}+1\\ &=1+\sum_{n=1}^\infty \left(1-5\right)\frac{1}{n5^n}\\ &=1-4\sum_{n=1}^\infty \frac{1}{n5^n}\\ &=1-4\sum_{n=1}^\infty \frac{(1/5)^n}{n}\\ &=1+4\ln \left(1-\frac{1}{5}\right)\\ &=1+4\ln (4/5) = 1-4\ln (5/4) \end{align}$$

Glen O
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