# The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4…$

Initially, I broke this series as $$1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$$, then I got two series as it is

$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$

After solving the first part, we reached at this point

$$= 1/20 - [1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$

It is given: $$\frac12\cdot \left(\frac15\right)^2+\frac23\cdot \left(\frac15\right)^3+\frac34\cdot \left(\frac15\right)^4+\cdots+\frac{n}{n+1}\cdot \left(\frac15\right)^{n+1}+\cdots$$ Consider the function: $$f(x)=\frac12\cdot x^2+\frac23\cdot x^3+\frac34\cdot x^4+\cdots+\frac{n}{n+1}\cdot x^{n+1}+\cdots$$ We want to find $$f\left(\frac15\right)$$. (Why?)

Take derivative from both sides: $$f'(x)=\color{blue}{x+2x^2+3x^3+\cdots +nx^n+\cdots}$$ Consider another function: $$g(x)=1+x+x^2+x^3+\cdots+x^n+\cdots=\frac1{1-x}, \quad \left(x=\frac15<1\right)$$ Take derivative from both sides: $$g'(x)=1+2x+3x^2+\cdots+nx^{n-1}+\cdots=\frac1{(1-x)^2}$$ Multiply both sides by $$x$$: $$xg'(x)=\color{blue}{x+2x^2+3x^3+\cdots+nx^n+\cdots}=\frac{x}{(1-x)^2}$$ Note that $$f'(x)=xg'(x)$$, so: $$f'(x)=\frac{x}{(1-x)^2}$$ Now we integrate both sides: $$\int f'(x)dx=\int \frac{x}{(1-x)^2}dx \Rightarrow\\ f(x)=\int \frac{x-1+1}{(1-x)^2}dx=-\int \frac1{1-x}dx+\int \frac1{(1-x)^2}dx =\\ \ln (1-x)+\frac1{1-x}+C$$ Note that $$f(0)=0$$: $$f(0)=\ln (1-0)+\frac1{1-0}+C=0 \Rightarrow C=-1$$ So: $$f(x)=\ln (1-x)+\frac1{1-x}-1$$ At last we plug $$x=\frac15$$: $$f\left(\frac15\right)=\ln \left(1-\frac15\right)+\frac1{1-\frac15}-1=\ln \frac45-\frac14.$$

• Thank you Friend. – Himanshu A Sep 11 at 6:14
• You are welcome! Good luck. – farruhota Sep 12 at 9:57

This second sum can be written as $$\sum_{n=2}^{\infty}\frac{1}{n}\left(\frac{1}{5}\right)^n = \frac{1}{25}\sum_{n=0}^{\infty}\frac{1}{n+2}\left(\frac{1}{5}\right)^n$$ This is the Lerch Transendent, so $$=\frac{1}{25}\phi\left(\frac{1}{5}, 1,2\right)$$ By the identities $$\phi(z,s,a) = \frac{1}{z}\left(\phi(z,s,a-1) - \frac{1}{((a-1)^2)^{s/2}}\right)$$, $$\frac{1}{z}\mathrm{Li}_n(z)=\phi(z,n,1)$$ and $$\mathrm{Li}_1(z)=-\ln(1-z)$$:

\begin{align} &=\frac{1}{25}\cdot5\left(\phi\left(\frac{1}{5},1,1\right)-1\right) \\ &= \frac{1}{5}\left(5\,\mathrm{Li}\left(\frac{1}{5}\right)-1\right) \\ &= -\ln\left(\frac{4}{5}\right)-\frac{1}{5}\end{align} Therefore, the final answer to your entire sum is \begin{align} &= \frac{1}{20} - \left(-\ln\left(\frac{4}{5}\right)-\frac{1}{5}\right)\\ &= \frac{1}{4}+\ln\left(\frac{4}{5}\right)\\ &\approx 0.026856\end{align}

More Reading: Lerch Transendent Lerch Identity Polylogarithm

• If the OP knew about the Lerch Transcendent, he would've already figured out the question. – Display name Sep 10 at 18:23
• I'll make a light effort to defend my upvote. The first sentence in this post is helpful to the OP. And the rest is helpful to the rest of the community. – Mason Sep 10 at 18:39
• @Displayname: I do not aware with Lerch Transendent. I was looking for 10+2 level simple solution. Anyways thanks for your effort. I have solved it just now. Will share my answer soon. – Himanshu A Sep 11 at 2:38

We know the GP progression formula for when common ratio less than $$1$$ which is pretty simple to derive and is $$\displaystyle \sum_{k=1}^\infty r^{k} = \dfrac{1}{1-r}$$ from there we can write the following, \begin{align} \int \sum\limits_{k=2}^\infty r^{k-1} dr = \int \frac{1}{1-r} dr \end{align}

After integration we get: \begin{align} \sum\limits_{k=2}^\infty \frac{r^{k}}{k}=-\ln(1-r)\end{align} Now start from $$k=2$$ instead of $$k=1$$ and let $$r=\dfrac{1}{5}$$.

• Hi Friend, Can you please share the complete solution along with answer. As I am looking to solve it using 10+2 level maths way. – Himanshu A Sep 10 at 17:37

Hint

For the time being, forget the $$\frac 15$$ and consider that what you need is $$S=\sum_{i=1}^\infty \frac i {i+1} x^{i+1}$$ Derive with respect to $$x$$ $$S'=\sum_{i=1}^\infty i x^{i}=x \sum_{i=1}^\infty i x^{i-1}=x\left(\sum_{i=1}^\infty x^{i}\right)'=x \times \left(\frac x {1-x}\right)'=\frac{x}{(1-x)^2}$$ Now, integration (by parts) to get $$S$$ and, when done, make $$x=\frac 15$$.

Thanks all who have given the answers. Some people use "Lerch Transenden" methods to solve some uses "integration", "convergence" etc. But I was looking to solve it as using 10+2 level math techniques.

Finally I got success in solving it. See my solution below -

= (1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4 +.........

= (1- 1/2)(1/5)^2 + (1- 1/3)(1/5)^3 + (1- 1/4)(1/5)^4 +.........

= (1/5)^2+(1/5)^3+(1/5)^4+.......−{1/2(1/5)^2+1/3(1/5)^3+1/4(1/5)^4+.....}

= (1/5)+(1/5)^2+(1/5)^3+(1/5)^4+.......−{(1/5)+1/2(1/5)^2+1/3(1/5)^3+1/4(1/5)^4+.....}

= (1/5)/(1- 1/5) - {-log(1- 1/5)}

= (1/5)/(4/5) + log(4/5)

= 1/4 + log(4/5)

Here log(1-x) = -x -x^2/2 -x^3/3 -..........

Once again thank you all of you.

• But how do you know that identity for $\log(1-x)$? Some people used integration just to obtain this identity. – Display name Sep 11 at 12:47
• Yes you have put a very valid concern. My answer is related with 10+2 level ALGEBRA math's chapter. When we studied series (AP,GP,HP.etc) we also studied "Exponential" & "Lograrthims" series. The formula above is clearly from "Lograrthims" series. So it's obvious for a student to know this. Thank you for raising a valid concern. – Himanshu A Sep 12 at 9:53