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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Where is $n$ in your sum? – Jan Eerland Dec 06 '15 at 14:08
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100 terms gives 3.751058201775122658119... No apparent pattern emerging as far as I can tell – Brevan Ellefsen Dec 12 '15 at 19:35
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What about 1000? I suspect it's going to be within $+0.1$ of your value. – Dec 12 '15 at 20:12
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@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. – Brevan Ellefsen Dec 12 '15 at 20:53
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Instead, use the upper bound that I showed in my answer, its complexity is far less than computing factorial. – Dec 12 '15 at 20:54
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@NimaBavari A wise idea. I'll calculate both if I can. – Brevan Ellefsen Dec 12 '15 at 20:55
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@Pls2 Your sum is an infinite (and convergent) series, not a partial sums. That's why it's independent of any variable. – Dec 12 '15 at 21:55
2 Answers
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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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@NimaBavari Mathematica estimates of this bound for $n$ terms are: $100: 44.55596381508841697933243577901742083327083106230589993948784728799415152911753...$ – Brevan Ellefsen Dec 12 '15 at 21:04
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$10$ terms: $44.555963815088416979332435779017420833270831062305899939448843541340243425608551$ – Brevan Ellefsen Dec 12 '15 at 21:06
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@NimaBavari the sum. It appears $10$ and $100$ terms differ very little, on the order of $3.9 \times 10^{-56}$ – Brevan Ellefsen 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.
$$\sum _{k=1}^{2} \frac{(k+1)^k k!}{k^2!}\approx 2.75$$ $$\sum _{k=1}^{10} \frac{(k+1)^k k!}{k^2!}\approx 2.75106$$ $$\sum _{k=1}^{50} \frac{(k+1)^k k!}{k^2!}\approx 2.75106$$
Michael Medvinsky
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