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I need to prove that the infinite series $$ \sum^\infty_{n=1} \frac{\textrm{sin}^2(n) + n^n}{1 + n^2e^nn!} $$ converges or diverges. I already showed that $n^n \leq n!e^n$ and $(1 + \frac{1}{n})^n \leq e$ $\forall n \in \mathbb{N}$ which must surely be useful to proof the divergence/convergence of this series but I don't know how. I am sure I must use the direct comparison test i.e. compare the sequence to some other sequence of which I know that its series diverges/converges. I tried $$ \frac{\textrm{sin}^2(n) + n^n}{1 + n^2e^nn!} \leq \frac{1+n^n}{1+n^2e^nn!} \leq \frac{1+n!e^n}{1+n^2e^nn!} \leq \frac{1+n!e^n}{n^2e^nn!} = \frac{1}{n^2}\left(\frac{1}{e^nn!} + 1\right) ... $$ and $$ \frac{\textrm{sin}^2(n) + n^n}{1 + n^2e^nn!} \geq \frac{n^n}{1 + n^2e^nn!} ... $$ but I don't know how I can simplify those expressions further to get a useful result.

Felix Gervasi
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  • So if each term looks like $ 1/ n^2$, then you know that the summation will be finite. We can't abound it from above by $1/n^2$, but could we do say $1/(n^2-1)$? Or $ 1/(n-1)^2$? – Calvin Lin Feb 29 '24 at 13:34
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    Hint: continue with your first inequalities and use $$\frac1{e^nn!}\le 1$$ – Kroki Feb 29 '24 at 13:36
  • Oh my... its embarrassing how long I looked at this inequality and didn't consider this. Is it possible that the limit is $\frac{\pi^2}{3}$? (not of this series of course but of the one I estimated with the inequalities) @Kroki – Felix Gervasi Feb 29 '24 at 13:47
  • Yes that's the value of that series – Kroki Feb 29 '24 at 13:53
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    @Kroki Okay and since that series converges the other one does too by the direct comparison test... thank you so much for your help! – Felix Gervasi Feb 29 '24 at 14:23

3 Answers3

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Of course if you know the Stirling equivalent $$n!\sim \big(\frac ne\big)^n\sqrt{2\pi n}$$ you have immediatly $u_n\sim{n^{-5/2}}$.

noradan
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Okay so I thought of answering my own question here in case anyone stumbles upon the same and summarize what was discussed in the comments above.
We want to know if the series $\sum^\infty_{n=1} \frac{\textrm{sin}^2(n)+n^n}{1+n^2e^nn!}$ converges or diverges. One way to do this is by using the direct comparison test. We can estimate the sequence of this series upwards as follows: $$ \frac{\textrm{sin}^2(n)+n^n}{1+n^2e^nn!} \leq \frac{1+n^n}{1+n^2e^nn!} \overset{*}{\leq} \frac{1+e^nn!}{1+n^2e^nn!} \leq \frac{1+e^nn!}{n^2e^nn!} = \frac{1}{n^2}\left(\frac{1}{e^nn!}+1\right) \leq \frac{2}{n^2}. $$ Since we know the limit of the series $$ \sum^\infty_{n=1} \frac{2}{n^2} = 2 \sum^\infty_{n=1} \frac{1}{n^2} = 2 \cdot \frac{\pi^2}{6} = \frac{\pi^2}{3} $$ we can conclude by the direct comparison test that $\sum^\infty_{n=1} \frac{\textrm{sin}^2(n)+n^n}{1+n^2e^nn!}$ must also converge.

* For this inequality we used that $n^n \leq n!e^n \ \forall n \in \mathbb{N}$. You can find the proof here for example.

Felix Gervasi
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We have $${\sin^2n+n^n\over 1+n^2e^nn!}\le {2\over n^2}{n^n\over e^nn!}$$ Let $a_n=\displaystyle{n^n\over e^nn!}.$ Then $$ {a_{n+1}\over a_n}=e^{-1}\left (1+{1\over n}\right )^n<1$$ Hence $a_n$ is decreasing, thus $a_n\le a_1=e^{-1}<2^{-1}.$ Therefore $${\sin^2n+n^n\over 1+n^2e^nn!}\le {1\over n^2}\le {1\over n-1}-{1\over n},\quad n\ge 2$$ Thus the series is convergent and its sum is less than or equal $1+1=2.$

Ryszard Szwarc
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