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I can establish that: $$\frac{n^n}{(2n+1)!}=\frac{n^n}{(2n)!(2n+1)}\le\frac{n^n}{n!}$$ But $$\sum_{n=1}^{+\infty}\frac{n^n}{n!}$$ diverges (by the ratio test). And even if it converged I wouldn't have info on the N that I need.

I might say that: $$\frac{n^n}{(2n+1)!}=\frac{n^n}{(2n)!(2n+1)}\le\frac{n^n}{2^n}=(\frac{n}{2})^n$$ which diverges as well.

I should be able to use the comparison test but I can't see how.

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Let $a_n=\frac{n^n}{(2n+1)!}$ then for $N,n\geq 1$, $$\frac{a_{n+1}}{a_n}=\frac{(1+1/n)^n}{2(2n+3)}< \frac{e}{2(2n+3)}<\frac{3}{10}\implies a_{N+n}< a_N \left(\frac{3}{10}\right)^{n}$$ Moreover let $S$ be the sum of the series then $$0<S-\sum_{n=1}^{N}\frac{n^n}{(2n+1)!}=\sum_{n=N+1}^{\infty}a_n< a_N\sum_{k=1}^{\infty}\left(\frac{3}{10}\right)^{k}.$$ Can you take it from here?

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  • $\begingroup$ Thanks for your answer. I'm probably missing a simple piece but I don't get why $\frac{a_{n+1}}{a_n}<\frac{3}{10}\implies a_{N+n}< a_N \left(\frac{3}{10}\right)^{n}$. Could you please clarify it? $\endgroup$ – zcb Apr 16 at 13:18
  • $\begingroup$ @zcb $a_{n+1}<3/10a_n$ implies that $a_{N+1}<3/10a_N$ and $a_{N+2}<3/10a_{N+1}$. Then $a_{N+2}<3/10a_{N+1}<(3/10)^2a_N$. The general case can be shown in a similar way. $\endgroup$ – Robert Z Apr 16 at 15:26
  • $\begingroup$ Thanks very much. It's clear now. $\endgroup$ – zcb Apr 16 at 15:36
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Note that\begin{align}\frac{\dfrac{(n+1)^{n+1}}{(2n+3)!}}{\dfrac{n^n}{(2n+1)!}}&=\frac{(n+1)^{n+1}}{n^n}\times\frac{(2n+1)!}{(2n+3)!}\\&=(n+1)\left(1+\frac1n\right)^n\times\frac1{(2n+2)(2n+3)}\\&=\frac12\left(1+\frac1n\right)^n\frac1{2n+3}\\&<\frac1{2n+3}\text{ since $\left(1+\frac1n\right)^n<e<4$}\\&<\frac12.\end{align}So,\begin{align}\left(\sum_{n=1}^\infty\frac{n^n}{(2n+1)!}\right)-\left(\sum_{n=1}^N\frac{n^n}{(2n+1)!}\right)&=\sum_{n=N+1}^\infty\frac{n^n}{(2n+1)!}\\&=\frac{(N+1)^{N+1}}{(2N+3)!}+\frac{(N+2)^{N+2}}{(2N+5)!}+\cdots\\&<\frac{(N+1)^{N+1}}{(2N+3)!}+\frac{(N+1)^{N+1}}{(2N+3)!}\times\frac12+\frac{(N+1)^{N+1}}{(2N+3)!}\times\frac1{2^2}+\cdots\\&=\frac{(N+1)^{N+1}}{(2N+3)!}\left(1+\frac12+\frac1{2^2}+\cdots\right)\\&=2\frac{(N+1)^{N+1}}{(2N+3)!}.\end{align}Since$$2\frac{4^4}{9!}=\frac1{2\,835}<\frac1{200},$$taking $N=3$ will be enough.

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  • $\begingroup$ Thanks for your answer. I can't understand why $\sum_{n=N+1}^\infty\frac{n^n}{(2n+1)!}<\frac{N^N}{(2N+1)!}\sum_{n=N+1}^\infty\frac1{2^n}$. Could you please clarify it? $\endgroup$ – zcb Apr 16 at 14:40
  • $\begingroup$ Actually, there was an error in what I wrote. I hope that everything is clear and correct now. $\endgroup$ – José Carlos Santos Apr 16 at 14:55
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Hint: try some stronger lower bound for denominator. For example, $(2n + 1)! = \prod\limits_{i=1}^{2n+1} i = \prod\limits_{i=1}^{n} i \cdot \prod\limits_{i=n+1}^{2n+1} i > 2^{n - 1} \cdot n^n$.

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