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Question:

Find a closed form for $\sum_{k=0}^{n}k^5$ using generating functions.

$Solution.$

We define $f(x)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}k^5x^n=\sum_{n=0}^{\infty}a_nx^n$.

Therefore, $$\begin{array}{l} a_{n} =\{1,1+32,1+32+243,1+32+243+1024,\dotsc \}\\ \\ b_{n} =\{1,32,243,1024,3125,7776,16807,32768,\dotsc \}\\ \\ c_{n} =\{1,31,211,781,2101,4651,9031,15961,\dotsc \}\\ \\ d_{n} =\{1,30,180,570,1320,2550,4380,6930,\dotsc \}\\ \\ e_{n} =\{1,29,150,390,750,1230,1830,2550,\dotsc \}\\ \\ f_{n} =\{1,28,121,240,360,480,600,720,\dotsc \}\\ \\ g_{n} =\{1,27,93,119,120,120,120,120,\dotsc \}\\ \\ h_{n} =\{1,26,66,26,1,0,0,0,\dotsc \}\\ \\ \Longrightarrow h_{n} =1+26x +66x^{2} +26x^{3} +x^{4}\\ \\ \Longrightarrow a_{n} =\frac{1+26x +66x^{2} +26x^{3} +x^{4}}{( 1-x)^{7}}\\ \\ \Longrightarrow f( x) =\frac{1+26x +66x^{2} +26x^{3} +x^{4}}{( 1-x)^{7}}\\ \\ =\frac{1}{( 1-x)^{7}} +\frac{26x}{( 1-x)^{7}} +\frac{66x^{2}}{( 1-x)^{7}} +\frac{26x^{3}}{( 1-x)^{7}} +\frac{x^{4}}{( 1-x)^{7}}\\ \\ \\ =\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n} +26\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n+1} +66\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n+2}\\ +26\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n+3} +\sum _{n=0}^{\infty }\binom{n+7-1}{7-1} x^{n+4}\\ \\ =\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n} +26\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n+1} +66\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n+2}\\ +26\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n+3} +\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n+4} \end{array}$$

By substitution, $$ \begin{array}{l} =\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n} +26\sum _{n=0}^{\infty }\binom{n-1+6}{6} x^{n} +66\sum _{n=0}^{\infty }\binom{n-2+6}{6} x^{n}\\ +26\sum _{n=0}^{\infty }\binom{n-3+6}{6} x^{n} +\sum _{n=0}^{\infty }\binom{n-4+6}{6} x^{n}\\ \\ =\sum _{n=0}^{\infty }\binom{n+6}{6} x^{n} +26\sum _{n=0}^{\infty }\binom{n+5}{6} x^{n} +66\sum _{n=0}^{\infty }\binom{n+4}{6} x^{n}\\ +26\sum _{n=0}^{\infty }\binom{n+3}{6} x^{n} +\sum _{n=0}^{\infty }\binom{n+2}{6} x^{n}\\ \\ =\sum _{n=0}^{\infty }\left[\binom{n+6}{6} +26\binom{n+5}{6} +66\binom{n+4}{6} +26\binom{n+3}{6} +\binom{n+2}{6}\right] x^{n}\\ \\ \Longrightarrow a_{n} =\binom{n+6}{6} +26\binom{n+5}{6} +66\binom{n+4}{6} +26\binom{n+3}{6} +\binom{n+2}{6} \end{array}$$

Now, I have checked over the Wolfram site, and it is incorrect. I don't know where I was wrong.

Chopin
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    Your original series is divergent. What is $n$ in $f(x)=a_n x^n$? What is $b_n$, $c_n$ and so on? – Gary Aug 28 '21 at 16:10
  • What is the source of this problem? Another question about sums of fifth powers appeared very recently. – lulu Aug 28 '21 at 16:11
  • @Gary you are right, I wrote this question really fast. My bad. Edited! – Chopin Aug 28 '21 at 16:15
  • @lulu I really don't know. – Chopin Aug 28 '21 at 16:16
  • It still does not make sense. The running index is $k$ in the first sum for $f$ but $n$ appears as the power of $x$. Also as I said the sum of $k^5$ up to infinity is a divergent sum. Isn't it $$ f(x) = \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 0}^n {k^5 } } \right)x^n } $$ what you want to write? – Gary Aug 28 '21 at 16:19
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    You really don't know what the source of your question is? – lulu Aug 28 '21 at 16:20
  • @Gary yes I have fixed that too. I really don't know what is going on with me today. I am really sorry! – Chopin Aug 28 '21 at 16:21
  • You still have $ \sum\nolimits_{k = 0}^\infty {k^5 } $ in two places, which is divergent. – Gary Aug 28 '21 at 16:22
  • It's called a "closed form", not "close form". – jjagmath Aug 28 '21 at 16:24
  • @Gary Right. Fixed that too... – Chopin Aug 28 '21 at 16:25
  • Please edit your post to include the source of this question. It is a duplicate of a recently asked question. – lulu Aug 28 '21 at 16:42
  • @lulu Can you provide the link that you are talking about? – Chopin Aug 28 '21 at 16:43
  • If you can not provide the source, the question should be closed. – lulu Aug 28 '21 at 16:43
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    @lulu What do you mean by providing the source? I do not have it, it is a part of self-practice. – Chopin Aug 28 '21 at 16:44
  • This MSE link may prove useful. – Marko Riedel Aug 28 '21 at 18:40
  • @MarkoRiedel in the title I have said the in this question you must derive the formula from generating functions. – Chopin Aug 29 '21 at 10:34
  • @lulu: OPs question in connection with his detailed reasoning looks sound to me. What was the reason to vote for closing this post? Kind regards, – Markus Scheuer Aug 30 '21 at 14:26
  • @MarkusScheuer A very similar question had appeared minutes prior to this one (find $\sum_1^n k^5$ without using Faulhaber). In these remote instruction days, we've had a lot of exam questions (and such) posted here. – lulu Aug 30 '21 at 15:45
  • @lulu: Ok, I see, thanks. Nevertheless this one differs clearly in quality by providing a derivation contrary to the other posts. But it's not that important for me ... – Markus Scheuer Aug 30 '21 at 16:20

1 Answers1

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It seems there's just an off-by-one error. Note that since \begin{align*} \color{blue}{f(x)}&=\sum_{n=0}^\infty\sum_{k=0}^n k^5 x^n=0+x+32x^2+276 x^3+1\,300 x^4\cdots\\ &=\color{blue}{\frac{x+26x^2+66x^3+26x^4+x^5}{(1-x)^7}} \end{align*} the coefficients $a_n$ should be considered as \begin{align*} \begin{array}{c|rrrrrr} n&0&1&2&3&4&\cdots\\ a_n&0&1&33&276&1\,300&\cdots\tag{1} \end{array} \end{align*}

Taking $a_n$ as in (1) will result in \begin{align*} \color{blue}{\sum_{k=1}^nk^5}&=\sum_{m=0}^4A(5,m)\binom{n+m+1}{6}\\ &\,\,\color{blue}{=\binom{n+1}{6}+26\binom{n+2}{6}+66\binom{n+3}{6}}\\ &\,\,\qquad\color{blue}{+26\binom{n+4}{6}+\binom{n+5}{6}} \end{align*} where $A(5,m): 1,26,66,26,1$ are Eulerian numbers which provide nice connections with $k$-th powers of natural numbers.

All other calculations of OP are correct. The sequence \begin{align*} &\binom{n+6}{6} +26\binom{n+5}{6} +66\binom{n+4}{6} \\ &\qquad+26\binom{n+3}{6} +\binom{n+2}{6}\qquad\qquad (n\geq 0) \end{align*} starts with $1,33,276,1\,300, \ldots$ corresponding to OPs start of the calculation.

Markus Scheuer
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