I want to compute $\displaystyle{\sum_{n=4}^{\infty}{1 \over n^{3} + n^{2}\cos\left(n\right)}.\quad}$ Can anyone help me or give a hint?
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2I don't see how. Converges, yes. If you had $\cos (\pi n)$ maybe something could be done – Will Jagy Sep 16 '16 at 01:48
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1What can we do with cos(pi n)? – Eric Zhao Sep 16 '16 at 02:03
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1And why is it converging? – Eric Zhao Sep 16 '16 at 02:05
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$\cos(\pi n)$ for integer $n$ is just $(-1)^n$ which is way easier to work with than the cosine of integer values. – rikhavshah Sep 16 '16 at 02:07
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2It converges since each term $\frac{1}{n^3+n^2cos(n)}$ is less than $\frac{1}{(n-1)^3}$, which converges by the p-series test, so by the direct comparison test your series converges. – rikhavshah Sep 16 '16 at 02:10
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but your fraction is different from mine – Eric Zhao Sep 16 '16 at 02:11
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oh i see, it still works by p test – Eric Zhao Sep 16 '16 at 02:11
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but how to compute it? – Eric Zhao Sep 16 '16 at 02:14
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Numerically, using a software package like Mathematica or Sage. – Math1000 Sep 16 '16 at 23:38
1 Answers
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This is not an answer but it is too long for an answer.
I do not think that a closed form of the summation could be found.
Using the same argument as user125261 in his/her comment, you can write $$\sum_{n=4}^{\infty}\frac{1}{n^3+n^2}\lt\sum_{n=4}^{\infty}\frac{1}{n^3+n^2\cos(n)}\lt \sum_{n=4}^{\infty}\frac{1}{n^3-n^2}$$ that is to say $$\frac{3 \pi ^2-29}{18}\lt\sum_{n=4}^{\infty}\frac{1}{n^3+n^2\cos(n)}\lt \frac{61-6 \pi ^2}{36}\approx 0.0495104$$ Now, computing partial sums, one could get the following results for $$S_p=\sum_{n=4}^{p}\frac{1}{n^3+n^2\cos(n)}$$ $$\left( \begin{array}{cc} p & S_p \\ 100 & 0.0419069 \\ 200 & 0.0419439 \\ 300 & 0.0419508 \\ 400 & 0.0419533 \\ 500 & 0.0419544 \\ 600 & 0.041955 \\ 700 & 0.0419553 \\ 800 & 0.0419556 \\ 900 & 0.0419558 \\ 1000 & 0.0419559 \end{array} \right)$$ which do not show a very fast convergence (this was done using Excel).
Claude Leibovici
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I tried it out on Python and for $p = 1000000$, $S_p = 0.04195636889897616$. Maybe for higher and higher values of $p$ it would just be around $S_p \approx 0.04196$
Edit: Tried for $S_0 = 10000000$ and $S_p$ still has the same answer, so either it converges to that number or it's just a limitation of Python. Either way I wouldn't try a higher value of $p$ as it might crash my computer.
– mopy Sep 16 '16 at 05:10 -
1@Aldon. Thanks for the info. I tried also and arrived to $\approx 0.041956368900332$ – Claude Leibovici Sep 16 '16 at 05:57