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Question:Prove that $ \sum_{n=1}^{\infty}{\frac{n^4}{5^n}}=\frac{285}{128}$

While doing questions on series and products,i got stuck in this question.I was not able to figure out any way how to prove this one.Convergence tests shows that this series converges but i don't know how to find its value.

Paras
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    One common tack in such problems is to start with $\sum_{n=1}^\infty r^n = \frac{r}{1-r}$, and then differentiate with respect to $r$. In this case, you'll have to differentiate four times. You may also have to combine lower derivatives in combination to arrive at $\sum_{n=1} n^4r^n$; be careful also with the lower limit of these sums. At the end, evaluate at $r = 1/5$ to arrive at your final result. – Brian Tung Jun 03 '20 at 20:12
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    For instance, taking the first derivative yields $\sum_{n=1}^\infty nr^{n-1} = \frac{d}{dr} \sum_{n=1}^\infty r^n = \frac{d}{dr} \frac{r}{1-r} = \frac{1}{(1-r)^2}$; then multiply both sides by $r$ to get $\sum_{n=1}^\infty nr^n = \frac{r}{(1-r)^2}$. – Brian Tung Jun 03 '20 at 20:15

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Here's a hint:

You should be able to evaluate the series $$\sum_{n=1}^\infty \frac{1}{5^n}$$ with a simple geometric series argument. More generally, this allows you to say something like $$\sum_{n=1}^\infty x^n = \frac{x}{1-x}$$ for $-1<x<1$. The magic happens when you take the derivative of this to get $$\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2},$$ which gives that $$\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2}.$$ However, we want a higher power of $x$. What happens if we take the derivative again? Can you generalize this enough to solve your problem?

Carl Schildkraut
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Equivalently, we want to evaluate $\left.\sum_nn^4x^n\right|_{x=1/5}$ with $\sum_n:=\sum_{n=\color{blue}{0}}^\infty$. By the binomial theorem, if $|x|<1$ then$$\begin{align}(1-x)^{-1}&=\sum_nx^n,\,\\(1-x)^{-2}&=\sum_n(n+1)x^n,\,\\(1-x)^{-3}&=\sum_n\tfrac12(n+1)(n+2)x^n,\,\\(1-x)^{-4}&=\sum_n\tfrac16(n+1)(n+2)(n+3)x^n,\,\\(1-x)^{-5}&=\sum_n\tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)x^n.\end{align}$$We'll find a linear combination by first getting the $n^4$ coefficient right, then $n^3$ etc. Since$$\begin{align}n^4&=24\cdot\tfrac{1}{24}(n+1)(n+2)(n+3)(n+4)\\&-60\cdot\tfrac16(n+1)(n+2)(n+3)\\&+50\cdot\tfrac12(n+1)(n+2)\\&-15\cdot(n+1)\\&+1,\end{align}$$we have$$\begin{align}\sum_nn^4x^n&=24(1-x)^{-5}-60(1-x)^{-4}\\&+50(1-x)^{-3}-15(1-x)^{-2}+(1-x)^{-1}.\end{align}$$Substituting $x=\tfrac15$ gives the desired result.

J.G.
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  • How did you arrive at the expansion for $x^4$? Some binomial identities? – Alex Jun 03 '20 at 22:08
  • @Alex Do you mean $n^4$? If so, the first line gets the $n^4$ term right, the first two lines get the $n^3$ term right etc. – J.G. Jun 03 '20 at 22:09