How to evaluate this series for $k > 1$? $$\sum_{n=1}^{\infty}\frac{1}{n k^n}$$

For $k = 2$, I tried to evaluate $\displaystyle \sum_{n = 0}^\infty \int_{1}^{2} x^{-(n+1)}dx = \int_{1}^{2} \sum_{n = 0}^\infty x^{-(n+1)}dx = \int_1^{2}\frac{1}{x(x-1)}dx$ $\displaystyle = \int_{1}^{2}\frac{1}{x-1}dx - \int_{1}^{2}\frac 1 x dx = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n2^n}\right)$

But both $\displaystyle \int_{1}^{2}\frac{1}{x-1}dx$ and $\displaystyle \sum_{n=1}^\infty\frac1n$ diverges. The answer is $\ln 2$, are these divergent terms equal?


$$\frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{nk^{n-1}} =\frac{1}{k} \sum_{n=1}^{\infty} \int^1_0 (x/k)^{n-1} dx=\frac{1}{k} \int^1_0 \sum_{n=1}^{\infty} (x/k)^{n-1} dx = \int^1_0 \frac{1}{k-x} dx= \log \left( \frac{k}{k-1} \right).$$


Hint: for $0<x<1$, we have $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$

Now integrate with respect to $x$, and then let $x=\frac{1}{k}$

  • $\begingroup$ integrate from where to where (limits)? $\endgroup$ – Santosh Linkha Mar 31 '13 at 11:14
  • $\begingroup$ Indefinite integral. What is the anti-derivative of $x^n$? $\endgroup$ – Glen O Mar 31 '13 at 11:15
  • $\begingroup$ what about the constant of integration? $\endgroup$ – Santosh Linkha Mar 31 '13 at 11:16
  • $\begingroup$ Think about it, and you'll figure it out. Remember that you want the resulting sum to look like the sum you're trying to evaluate. $\endgroup$ – Glen O Mar 31 '13 at 11:17
  • $\begingroup$ if $0<x<1$ then it must be definite integral from $0$ to $1$, else if it were indefinite integral we would get constant of integration. Thanks for response!! (+1) $\endgroup$ – Santosh Linkha Mar 31 '13 at 11:30

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