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Find the sum of the series $$ \sum \limits_{1}^{\infty} \frac{1}{n. 5^{n}} $$ $$$$ I have tried in this way, the $ n^{th}$ term of the series is $t_{n}=\frac{1}{n.5^{n}}$. Then $ S_{n}=t_{1}+t_{2}+.......+t_{n}$ But i can't proceed further , please anyone help me, thanks.

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  • $\begingroup$ The goal of the exercise is most likely checking if you know by heart the series previously covered in class/in the book. So, I think there is a good chanche you can find the formula for $\sum_{k=1}^\infty \frac{x^n}n$ somewhere in your notes. $\endgroup$ – user228113 Mar 3 '17 at 11:37
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Hint

Consider the function:

\begin{align} f_n(x) = \sum\limits _{k=1}^n x^{k-1} =\frac{1-x^{n}}{1-x} \end{align} Take the limits and integrate both sides and remember that the geometric series converges uniformly on interval $[a,b]$ provided that $-1<a<b<1$, so we get: \begin{align}\lim\limits_{n \to \infty} \int f_n(x) dx=\lim\limits_{n \to \infty}\int \sum\limits _{k=1}^n x^{k-1} dx=\lim\limits_{n \to \infty} \int \frac{1-x^{n}}{1-x} dx\end{align} We can take the limit inside the integral because of uniform convergence. \begin{align} \int \lim\limits_{n \to \infty} \sum\limits _{k=1}^n x^{k-1} dx = \int \sum\limits_{k=1}^\infty x^{k-1} dx = \int \lim\limits_{n \to \infty} \frac{1-x^{n}}{1-x} dx = \int \frac{1}{1-x} dx \end{align} Now we get after integrating: \begin{align} \sum\limits_{k=1}^\infty \frac{x^{k}}{k}=-\ln(1-x)+C\end{align} You get a integration constant $C$ so put $x=0$ and solve for $C$. Put $x=1/5$ and your are done.

I made some edits to make it more understandable, I hope this helps.

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This problem can be solved by expressing $1/n$ as $\int^1_0 x^{n-1} dx$

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  • $\begingroup$ how to find the value $\endgroup$ – mmath Mar 3 '17 at 12:06
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HINT:$$-\ln(1-u)=\sum_{r=1}^\infty\dfrac{u^r}r$$

$$\dfrac1{n\cdot5^n}=\dfrac{\left(\dfrac15\right)^n}n$$

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