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I was solving this problem in which I need to find the sum :-
$\frac{3}{5}+\frac{6}{5^2}+\frac{11}{5^3}+\frac{18}{5^4}+.....$

And I was able to derive the general form of any $n^{th}$ term of this series which is $\frac{n^2+2}{5^n}$ but I am not able to think how can we proceed from here. How can we find the sum from the general form that I have derived? Please help me on this !!!

Thanks in advance !!!

Ganit
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  • Hint : If you know how to differentiate, try to differentiate $\sum_{n}x^n$ twice – Infinity_hunter Oct 06 '21 at 12:07
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    Similar questions: https://math.stackexchange.com/q/2209638/42969, https://math.stackexchange.com/q/1955505/42969, https://math.stackexchange.com/q/200658/42969, https://math.stackexchange.com/q/3992229/42969 – Martin R Oct 06 '21 at 12:08
  • I know how to differentiate but I am not able to get how can we use this here? How did you get to what needs to be differentiated? Can you please put down your solution so that I can get to know about your way of solving this? That will be really helpful. – Ganit Oct 06 '21 at 12:09
  • here is a very similar one – Infinity_hunter Oct 06 '21 at 12:14
  • @Ganit . Use the fact that $\frac{1}{1-x} = \sum_{r=0}^{\infty}x^{r},,,-1<x<1$. You can derive this using binomial theorem for negative index. Formally it is the maclaurin expansion of $\frac{1}{1-x}$. Now you can differentiate it once. Then multiply by x on both sides and differentiate it again. then substitute the value of $x=\frac{1}{5}$ – Mr.Gandalf Sauron Oct 06 '21 at 12:16
  • Use $(n^2+2)x^n=[n(n-1)+n+2] x^n=n(n-1)x^n+n x^n +2 x^n$ – Claude Leibovici Oct 06 '21 at 12:31

2 Answers2

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$$\frac{1}{1-x}=\sum_{r=0}^{\infty}x^{r}$$

Differentiate once:-

$$\frac{1}{(1-x)^{2}}=\sum_{r=0}^{\infty}rx^{r-1}$$

Multiply by $x$ on both sides:-

$$\frac{x}{(1-x)^{2}}=\sum_{r=0}^{\infty}rx^{r}$$

Differentiate again:-

$$\frac{1}{(1-x)^{2}} +\frac{2x}{(1-x)^{3}}=\sum_{r=0}^{\infty}r^{2}x^{r-1}$$

Multiply by $x$ again:-

$$\frac{x}{1-x} +\frac{2x^{2}}{(1-x)^{3}}=\sum_{r=0}^{\infty}r^{2}x^{r}$$

Substitute $x=\frac{1}{5}$ in LHS to get $\sum_{n=0}^{\infty}\frac{n^{2}}{5^{n}}$

And I think you can manage the $2\sum_{n=0}^{\infty}\frac{1}{5^{n}}=\frac{2}{1-\frac{1}{5}}=\frac{5}{2}$ using the geometric series formula.

Note that all of these is only valid within the radius of convergence of the series.

So all of these only make sense when $-1<x<1$. Otherwise all of what we did was absolute nonsense.

Mr.Gandalf Sauron
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  • " Otherwise all of what we did was absolute nonsense.", not so fast : https://en.wikipedia.org/wiki/Ces%C3%A0ro_summation – jimjim Oct 06 '21 at 12:31
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    The op is looking for the closed form. He has already assumed convergence. So substituting $x=\frac{1}{5}$ in the expressions in LHS and RHS becomes non-sense – Mr.Gandalf Sauron Oct 06 '21 at 14:13
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If you have the general term for the sequence, it is pretty straightforward to proceed.

$T_n$ = $ \frac {n^2+2}{5^n}$

We let $S= \sum T_n$

$S= \frac {3}{5} +\frac{6}{5^2} +\cdots$

Now, we multiply the whole sum by $\frac {1}{5}$ , and subtract it from S

$ \frac{4S}{5}$=$ \sum \frac {2n+1}{5^{n+1}}$ =$\frac{3}{5} + \frac{3}{5^2}+\frac {5}{5^3} +\cdots$

We repeat this again , to obtain a geometric series.

So, we get

$\frac{16S}{25}$ = $\frac{3}{5} +\frac{2}{5^3} +\frac{2}{5^4}+\cdots $

This is a geometric sum with a term extra, the $\frac{3}{5}$.

$\frac{16S}{25}$ =$\frac{3}{5} + \frac{2}{5^3} ( 1+\frac{1}{5} +\cdots )$

This gives us $ \frac{16S}{25} = \frac{3}{5} + \frac{1}{50} $

$\Rightarrow$ $S$=$\frac{31}{32}$

Anuj jha
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