JEE Series Summation

$\displaystyle\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ Now differentiate the series (using the quotient rule - see bottom) and multiply by ${x}$ to obtain $\displaystyle\sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2}$ Now repeat the previous step to obtain: $\displaystyle\sum_{n=0}^{\infty} n^2 x^n = \frac{x(1+x)}{(1-x)^3}$ Now let x = $\frac{1}{2}$ $\large{\therefore\displaystyle\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6}$ Quotient rule: $y = \frac{u}{v} \implies\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

Solution "without using Calculus" : Let's denote $S_{k}=\sum_{n=1}^{\infty} {\frac{n^{k}}{2^{n}}}$ (see the remark at the end of the solution) . Since $S_{0}$ is the sum of a geometric series whose first term is $\frac{1}{2}$ and ratio $\frac{1}{2}$ , then $S_{0}=\frac{1}{2} \frac{1}{1 - \frac{1}{2}}=1.$ Now, we can see that $S_{1}=\sum_{n=1}^{\infty}{\frac{n}{2^{n}}}=\frac{1}{2}+ \sum_{n=2}^{\infty}{\frac{n}{2^{n}}}=\frac{1}{2}+\sum_{n=1}^{\infty}{\frac{n+1}{2^{n+1}}}=$ $=\frac{1}{2}+\frac{1}{2}\sum_{n=1}^{\infty}{\frac{n+1}{2^{n}}}=\frac{1}{2}+\frac{1}{2} (S_{1}+S_{0})$ $=\frac{1}{2}+\frac{1}{2} (S_{1}+ 1)$ Solving the resulting equation for $S_{1}$ we obtain that $S_{1}=2.$ In a similar way $S_{2}=\frac{1}{2} +\sum_{n=1}^{\infty}{\frac{(n+1)^{2}}{2^{n+1}}}=$ $= \frac{1}{2} +\sum_{n=1}^{\infty}{\frac{n^{2}}{2^{n+1}}}+\sum_{n=1}^{\infty}{\frac{2n}{2^{n+1}}}+\sum_{n=1}^{\infty}{\frac{1}{2^{n+1}}}=$ $= \frac{1}{2} + \frac{1}{2} S_{2}+S_{1}+ \frac{1}{2} S_{0}=3+\frac{1}{2} S_{2}.$ Solving the resultant equation we get that $S_{2}=6$ .

Apparently we have not used Calculus, but for this algebraic method to work the series with sums $S_{0}$ , $S_{1}$ and $S_{2}$ must be convergent, and the corresponding proof requires the use of Calculus methods.

Remark : Notice that the term of the series corresponding to $n=0$ is zero, and that is why we assume that the first value of $n$ in the summation is 1 .

$\begin{aligned} S & = \sum_{n=\color{#3D99F6}{0}}^\infty \frac{n^2}{2^n} \\ & = \sum_{n=\color{#D61F06}{1}}^\infty \frac{n^2}{2^n} \\ & = \sum_{n=0}^\infty \frac{(n+1)^2}{2^{n+1}} \\ & = \sum_{n=0}^\infty \frac{n^2+2n+1}{2^{n+1}} \\ & = \frac{1}{2} \sum_{n=0}^\infty \frac{n^2}{2^n} + \sum_{n=0}^\infty \frac{n}{2^n} + \frac{1}{2} \sum_{n=0}^\infty \frac{1}{2^n} \\ & = \frac{1}{2}S + \sum_{n=0}^\infty \left. x \frac{d}{dx} x^n \right|_{x=\frac{1}{2}} + \frac{1}{2} (2) \\ \implies \frac{1}{2}S & = \left. x \frac{d}{dx} \sum_{n=0}^\infty x^n \right|_{x=\frac{1}{2}} + 1 \\ & = \left. x \frac{d}{dx} \frac{1}{1-x} \right|_{x=\frac{1}{2}} + 1 \\ & = \left. \frac{x}{(1-x)^2} \right|_{x=\frac{1}{2}} + 1 \\ & = 2 + 1 \\ \implies S & = \boxed{6} \end{aligned}$

Non calculus solution:

$S=x+4{ x }^{ 2 }+9{ x }^{ 3 }+16{ x }^{ 4 }...$

Then $Sx={ x }^{ 2 }+4{ x }^{ 3 }+9{ x }^{ 4 }+16{ x }^{ 5 }...$

So $S-Sx=x+3{ x }^{ 2 }+5{ x }^{ 3 }+7{ x }^{ 3 }$

Now, consider $x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }...=P\\ P-\frac { P }{ x } =-1\\ P=\frac { -1 }{ 1-\frac { 1 }{ x } }$

So, $S(1-x)-P={ 2x }^{ 2 }+{ 4x }^{ 3 }+6{ x }^{ 4 }...=2\left( { x }^{ 2 }+2{ x }^{ 3 }+3{ x }^{ 4 }... \right) \\$

So now, $\\ { x }^{ 2 }+2{ x }^{ 3 }+3{ x }^{ 4 }...=a\\ ax={ x }^{ 3 }+2{ x }^{ 4 }+3{ x }^{ 5 }...\\ a-ax={ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 }+{ x }^{ 5 }...=P-x$

Therefore, $S(1-x)-P=2a=\frac { 2\left( P-x \right) }{ 1-x }$

$S=\frac { \frac { 2\left( P-x \right) }{ 1-x } +P }{ 1-x } =\frac { \frac { 2\left( \frac { -1 }{ 1-\frac { 1 }{ x } } -x \right) }{ 1-x } +\frac { -1 }{ 1-\frac { 1 }{ x } } }{ 1-x }$

If you are too lazy to simplify like me, just sub $x=0.5$ to get $6$ . Bye

Logic behind:

To show the logic I used, I'm just going to give a quick clue:

${ n }^{ 2 }-{ \left( n-1 \right) }^{ 2 }=2n-1\Longrightarrow S-Sx\\ 2n-1+1=2n\Longrightarrow S-Sx-P\\ 2n-(2n-1)=1\Longrightarrow a-ax=P-x$

Hint:: Try to genralise : $\large{\displaystyle{\sum _{ n=0 }^{ \infty }{ { n }^{ 2 }{ x }^{ n } } =\cfrac { x(x+1) }{ { (1-x) }^{ 3 } } }}$