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How to find the sum of the following summation?

$$\sum_{n=0}^\infty n^2 x^n $$

I was told to split it into two summations, but I don't see any help in that.

(original problem image)

Blue
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    Yeah I'm not sure what you might gain from that or how it might be meaningfully done. On the premise $|x|<1$, I would note

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

    and then take the derivative of both sides, multiply both sides by $x$, take another derivative, and multiply by $x$ again.

     – PrincessEev May 15 '20 at 21:04
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    That's precisely where splitting the sum into two helps, you get $x^2\sum n(n-1)x^{n-2}+x\sum nx^{n-1}$. – Isaac Ren May 15 '20 at 21:05
  • @IsaacRen How can I continue from this point and why did you split them into this in specific? there are trillions of options –  May 15 '20 at 21:07
  • Welcome to Math.SE! The community here prefers/expects questions to include something of what the asker knows about the problem. Even if you don't know exactly what to do, you can provide context that helps answerers tailor their responses to best serve you, without wasting time (theirs or yours) explaining what you already know or using techniques above your skill level. For instance, @EeveeTrainer has suggested taking a derivative, which is a calculus-level strategy; is this allowed? Or are you looking for a pre-calculus-based approach? The more you can say, the better. – Blue May 15 '20 at 21:14
  • The derivative of $x^n$ is $nx^{n-1}$, and the derivative of that is $n(n-1)x^{n-2}$, and their two coefficients $n$ and $n(n-1)$ sum to $n^2$. – anon May 15 '20 at 21:14

1 Answers1

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A few people have given elements of a solution in the comments, but for the sake of clarity I'll write it out here. As @Eevee Trainer said, we have the equality $$\sum_{n=0}^\infty x^n=\frac 1{1-x}.$$ As @runway44 suggested, differentiating this equation once gives $$\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}.$$ Differentiating it again, we get $$\sum_{n=0}^\infty n(n-1)x^{n-2}=\frac{2}{(1-x)^3}.$$ This is why I suggested the split $$\sum_{n=0}^\infty n^2x^n=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+x\sum_{n=0}^\infty nx^{n-1}.$$ Using the previous calculations, we conclude that $$\sum_{n=0}^\infty n^2x^n=\frac{2x^2}{(1-x)^3}+\frac{x}{(1-x)^2}=\frac{x^2+x}{(1-x)^3}.$$

Isaac Ren
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  • Thanks, I never though of deriving! A new way of solution :-) –  May 15 '20 at 21:28
  • yep, it's a very useful tool for power series like these :) – Isaac Ren May 15 '20 at 21:30
  • what if I was asked to calculate sigma of (n+2)*X^(n)? I did integral twice and reached (x^(n+2))/(n+1) But still I can't calculate sigma for this one –  May 15 '20 at 21:50
  • You always want to get something of the form $nx^{n-1}$. For $\sum(n+2)x^n$, you need to divide by $x$: $\sum(n+2)x^n=\frac 1x\sum(n+2)x^{n+1}=\frac 1x\sum_{n=2}^\infty nx^{n-1}$. – Isaac Ren May 15 '20 at 22:01
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    @JohnOmielan Of course! I got mixed up with the French vocabulary. – Isaac Ren May 15 '20 at 22:03
  • so I got that it's equal to 1/ ((x)*(1-x)^2) But I got wrong answer (my loop starts from n=0) –  May 15 '20 at 22:11
  • $\sum_{n=2}^\infty nx^{n-1}=\frac{1}{(1-x)^2}-x$, so it should be $\frac{1}{x(x-1)^2}-1$. – Isaac Ren May 15 '20 at 22:43