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How do I prove convergence or divergence of this series? I can't prove that it diverges using the divergence test (sequence converges to zero). Thanks in advance!

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  • $\begingroup$ Have you tried comparison test? $\endgroup$ – Tortuga Feb 17 '19 at 19:11
  • $\begingroup$ Do you know about the limit comparison test? $\endgroup$ – Mark Feb 17 '19 at 19:11
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    $\begingroup$ You cant prove that it diverges, because it converges. $\endgroup$ – babemcnuggets Feb 17 '19 at 19:12
  • $\begingroup$ Wait is that always true babemcnuggets? I didn't think getting 0 when taking the limit proved convergence $\endgroup$ – J.W. Feb 17 '19 at 19:15
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    $\begingroup$ It doesn't - for example take $1/n$, but the series here converges, so one cannot prove that it diverges as you stated. $\endgroup$ – Peter Foreman Feb 17 '19 at 19:17
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As $$\frac{1}{n^4+n+1}\lt\frac{1}{n^4}$$ for all $n\ge1$, we can say that $$\sum_{n=1}^\infty \frac{1}{n^4+n+1}\lt \sum_{n=1}^\infty \frac{1}{n^4}=\zeta{(4)}=\frac{\pi^4}{90}$$ $$\therefore\sum_{n=1}^\infty \frac{1}{n^4+n+1}\lt \frac{\pi^4}{90}$$ So the series converges.

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  • $\begingroup$ haha, the zeta function is neat. But I don't think the OP know why $\sum\frac{1}{n^4}=\zeta(4)$. Probably a geometric test can help the OP more or less :) $\endgroup$ – Tortuga Feb 17 '19 at 19:18
  • $\begingroup$ I mean I've never been taught about testing for convergence in school either... $\endgroup$ – Peter Foreman Feb 17 '19 at 19:20
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There are lots of different ways, but the simplest is the use of a simple inequality using the harmonic series:

$$\sum_{n=1}^{\infty} \frac{1}{n} > \sum_{n=1}^{\infty} \frac{1}{n^4+n+1} \space \forall n\geq 1$$

and the harmonic series is famously the largest series of the form to diverge, such that any series that fulfills the following criteria converges: $$\sum_{n=1}^{\infty} \frac{1}{n^a},a > 1$$

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Since the series with 1/n^4 converges, the series with 1/(n^4+n+1) also does, because since the denominator is bigger, the fraction is smaller.

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We have $n^4+n+1\gt n^2+n$ for all $n\ge1$, so we can bound the series with a telescoping sum:

$$\sum_{n=1}^\infty{1\over n^4+n+1}\lt\sum_{n=1}^\infty{1\over n^2+n}=\sum_{n=1}^\infty\left({1\over n}-{1\over n+1} \right)=\left(1-{1\over2} \right)+\left({1\over2}-{1\over3}\right)+\left({1\over3}-{1\over4}\right)+\cdots=1$$

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