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I want to prove that $$\sum_{n=1}^\infty{n^2a^{n-1}}=\frac{1+a}{(1-a)^3}$$

I start off at the sum and try to work my way into the equation.

I know that the sum is:

$$σ_{n}=1+2^2+a+3^2a^2+4^2a^3+\dots+n^2a^{n-1} (1)\Leftrightarrow$$ $$aσ_{n}=a+2^2a^2+3^2a^3+\dots+n^2a^n (2)$$

If I subtract (2) from (1), I get: $$(1-a)σ_{n}=1+2^2a+3^2a^2+\dots+n^2a^{n-1}-a-2^2a^2-\dots-n^2a^n \Leftrightarrow$$

$$(1-a)σ_{n}=(n^2+(n-1)^2)a^{n-1}-n^2a^n \Leftrightarrow$$ $$σ_{n}=\frac{n^2-(n-1)^2-n^2a^n}{1-a} \Leftrightarrow$$ $$σ_{n}=-\frac{1+n^2a^n}{1-a}$$

and that is what I've got so far. How can I continue solving this?

I don't know if this is the best solution of the proof, but my textbook had a similar proof for $$\sum_{n=1}^{\infty}na^{n-1}=\frac{1+a}{(1-a)^2}$$

So, I tried using the same way.

Shevliaskovic
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2 Answers2

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$\textbf{hint}$ $$ \sum n\dfrac{d}{da}a^n =\frac{d}{da}\sum na^n= \sum n^2a^{n-1} $$ Do you need me to continue?

Chinny84
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  • what is that $d$ ? Is it the derivative? – Shevliaskovic Jan 09 '15 at 18:17
  • Aren't you treating $a$ as a variable when it is a constant? – AlonAlon Jan 09 '15 at 18:18
  • I am treating it as a variable similar to differentiating under the integral sign. – Chinny84 Jan 09 '15 at 18:18
  • Is there any way to solve this without using integrals? I know how to use them, but the Series chapter in my book is before the integrals chapter. (The question above is from the series chapter). – Shevliaskovic Jan 09 '15 at 18:20
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    @Shevliaskovic For $|a|<1$, the classic geometric series is

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

    Chinny84's Hint is to take the derivative with respect to $a$ on both sides, twice. There are no integrals.

     – Zubin Mukerjee Jan 09 '15 at 18:21
  • @Chinny84 yup, I know that. I also know that $$\sum_{n=1}^{\infty}na^{n-1}=\frac{1+a}{(1-a)^2}$$ – Shevliaskovic Jan 09 '15 at 18:22
  • It is not an integral .. You can solve using series. Differentiating under the integral sign was just showing you another area where you have a constant and treat it as variables – Chinny84 Jan 09 '15 at 18:22
  • Then to finish off then make my hint similar to yours :). – Chinny84 Jan 09 '15 at 18:22
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We have to check that:

$$\left(1-3a+3a^2-a^3\right)\sum_{n=0}^{+\infty}(n+1)^2 a^n = 1+a. \tag{1}$$ Let $\delta$ be the backward difference operator: $$ \delta g(n) \triangleq g(n)-g(n-1).\tag{2}$$ If we set $b_n=(n+1)^2\cdot I_{n\geq 0}$, line $(1)$ can be written as: $$ \sum_{n=1}^{+\infty} \delta^3(b_n)\, a^n = 1+a. \tag{3} $$ For any $n\geq 3$, we have $\delta^3(b_n)=0$, since if $g$ is a polynomial, the degree of $\delta g$ is just the degree of $g$ minus one. So we just have to check that: $$ \delta^3(b_0)=1,\quad \delta^3(b_1)=1,\quad \delta^3(b_2)=0. \tag{4}$$ This is easy to do: $$(1,4,9,16,\ldots)\stackrel{\delta}{\longrightarrow}(1,3,5,7,\ldots)\stackrel{\delta}{\longrightarrow}(1,2,2,2,\ldots)\stackrel{\delta}{\longrightarrow}(1,1,0,0,\ldots).\tag{5}$$

Jack D'Aurizio
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