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I have to find sum (or the least upper bound) of infinite series (exact expression not decimal number) of series $\begin{equation} \sum_{k=1}^{+\infty}\frac{k!(1+k)^k}{(k^2)!} \end{equation}$. I am clueless, thank you for any help.

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  • $\begingroup$ Where is $n$ in your sum? $\endgroup$ – Jan Dec 6 '15 at 14:08
  • $\begingroup$ 100 terms gives 3.751058201775122658119... No apparent pattern emerging as far as I can tell $\endgroup$ – Brevan Ellefsen Dec 12 '15 at 19:35
  • $\begingroup$ What about 1000? I suspect it's going to be within $+0.1$ of your value. $\endgroup$ – user98186 Dec 12 '15 at 20:12
  • $\begingroup$ @NimaBavari I'll write a program to calculate it. I'll need the program to cancel out as much of the denominator as possible... $(1000^2)!$ is a pretty large number for standard floating point arithmetic. $\endgroup$ – Brevan Ellefsen Dec 12 '15 at 20:53
  • $\begingroup$ Instead, use the upper bound that I showed in my answer, its complexity is far less than computing factorial. $\endgroup$ – user98186 Dec 12 '15 at 20:54
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Stirling formula gives

$$\frac {k! (1 + k)^k} {(k^2)!} < k^{2k - k^2 - \frac {1} {2}} e^{k^2 - k + 1 + \frac {k - 1} {12 k^2}}.$$

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  • $\begingroup$ This is very nice! $\endgroup$ – GGG Dec 12 '15 at 20:18
  • $\begingroup$ @user276387 Thanks a lot! $\endgroup$ – user98186 Dec 12 '15 at 20:27
  • $\begingroup$ @NimaBavari Mathematica estimates of this bound for $n$ terms are: $100: 44.55596381508841697933243577901742083327083106230589993948784728799415152911753...$ $\endgroup$ – Brevan Ellefsen Dec 12 '15 at 21:04
  • $\begingroup$ $10$ terms: $44.555963815088416979332435779017420833270831062305899939448843541340243425608551$ $\endgroup$ – Brevan Ellefsen Dec 12 '15 at 21:06
  • $\begingroup$ @BrevanEllefsen the sum or the sequence? $\endgroup$ – user98186 Dec 12 '15 at 21:08
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These series $\sum_n a_n$ has a general term $a_n$ that decay extremely fast for example

$a_1=2$

$a_1=0.75$

$a_3=0.0010582$

$a_4=7.16922\times10^{-10}$

$a_5=6.01578\times10^{-20}$

$a_6=2.27712\times10^{-34}$

Therefore the main contribution to the sum is by the first terms.

$$\sum _{k=1}^{2} \frac{(k+1)^k k!}{k^2!}\approx 2.75$$ $$\sum _{k=1}^{10} \frac{(k+1)^k k!}{k^2!}\approx 2.75106$$ $$\sum _{k=1}^{50} \frac{(k+1)^k k!}{k^2!}\approx 2.75106$$

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