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How would you prove convergence/divergence of the following series?

$$\sum_{n\ge0} \sin (\pi \sqrt{n^2+n+1}) $$

I'm interested in more ways of proving convergence/divergence for this series.

My thoughts

Let $$u_{n}= \sin (\pi \sqrt{n^2+n+1})$$ trying to bound $$|u_n|\leq |\sin(\pi(n+1) )| $$ since $n^2+n+1\leq n^2+2n+1$ and $\sin$ is decreasing in $(0,\dfrac{\pi}{2} )$

$$\sum_{n\ge0}|u_n|\leq \sum_{n\ge0}|\sin(\pi(n+1) )|$$ or $|\sin(\pi(n+1) )|=0\quad \forall n\in \mathbb{N}$ then $\sum_{n\ge0}|\sin(\pi(n+1) )|=0$ thus $\sum_{n\ge0} u_n$ is converge absolutely then is converget

any help would be appreciated

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    $\begingroup$ I think your bound is not correct because sinus is not a non-decreasing function. $\endgroup$ – Alex Silva Jan 1 '15 at 16:36
  • $\begingroup$ is decreasing in $(0,\dfrac{\pi}{2} )$ $\endgroup$ – Educ Jan 1 '15 at 16:37
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    $\begingroup$ But $\pi\sqrt{n^2+n+1}$ does not belong to $(0,\pi/2)$. More to the point, a prerequisite for the series to converge would be to have $u_n\xrightarrow[n\to\infty]{} 0$. Is it the case? $\endgroup$ – Clement C. Jan 1 '15 at 16:39
  • $\begingroup$ @ClementC. i don't think so, it is diverge $\endgroup$ – Educ Jan 1 '15 at 16:42
  • $\begingroup$ Then, if $u_n$ does not converge to 0, the series $\sum u_n$ cannot converge. $\endgroup$ – Clement C. Jan 1 '15 at 16:43
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Hint: A necessary condition for the convergence of an infinite series $\sum\limits_{n=0}^\infty a_n$ is that the limit $\lim\limits_{n \to \infty} a_n$ should exist and be equal $0$. So for your series investigate the following limit :

$$ \lim_{n\to\infty}\sin\left(\pi\sqrt{n^2+n+1}\right). $$

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  • $\begingroup$ Why? Isn't $\sqrt{n^2+n+1} \approx \sqrt{n^2+2n+1} = (n+1)$ as $n \rightarrow \infty$ ? So $\sin(\pi\sqrt{n^2+n+1}) \rightarrow 0$ $\endgroup$ – Vladimir Fomenko Jan 1 '15 at 16:42
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    $\begingroup$ @Educ Then you'll probably upvote not that many answers. Many members in this site think that hints and not completely detailed answers is what serves askers the best. In fact, completely detailed answers are more usually given by newcomers or greenhorns than by veterans. $\endgroup$ – Timbuc Jan 1 '15 at 17:11
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    $\begingroup$ @Alex Silva I have a bit of egg on my face. Initially I assumed that the limit that I proposed earlier converged to zero, but upon further investigation I realize that that was very incorrect. In fact, $$\lim_{n\to\infty}\bigg((n+1)-\sqrt{n^2+n+1}\bigg) = \frac{1}{2}$$ Kudos to you for your insight. $\endgroup$ – John Joy Jan 1 '15 at 17:29
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    $\begingroup$ @Educ, you can distinguish between "more usually" and "always", right? $\endgroup$ – Timbuc Jan 1 '15 at 17:36
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    $\begingroup$ @Educ Perhaps I can, perhaps I cannot, yet what I wrote, and this is what you addressed, is "usually" . :) $\endgroup$ – Timbuc Jan 1 '15 at 18:43
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$$\begin{align}\lim_{n\to\infty} \sin(\pi\sqrt{n^2+n+1}) &= \lim_{n\to\infty} \sin\Bigg(\pi \ \sqrt{(n+\frac{1}{2})^2 - \frac{1}{4} + 1}\Bigg) \\&= \lim_{n\to\infty} \sin\Bigg(\pi \ \sqrt{(n+\frac{1}{2})^2 + \frac{3}{4}\Big)}\Bigg) \\&= \lim_{n\to\infty} \sin\Bigg(\pi \ \Big(n+\frac{1}{2}\Big)\sqrt{1 + \frac{3}{4(n+\frac{1}{2})^2}}\Bigg)\end{align}$$

then this limit changes between $1$ and $-1$.

Thus you may conclude that the series diverges.

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    $\begingroup$ I think this is very nice and correct, since expression inside the squared root is almost 1 for big n's and what is left is $\;\sin\pi(n+0.5)\;$ . $\endgroup$ – user177692 Jan 1 '15 at 18:09
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$$\lim_{n\rightarrow\infty}{\sin\left(\pi\cdot\sqrt{n^2+n+1}\right)}=\sin{\lim_{n\rightarrow\infty}\left(\pi\cdot\sqrt{n^2+n+1}\right)}=$$

$$=\sin\left(\pi\cdot\lim_{n\rightarrow\infty}\left(\sqrt{n^2+n+1}-(n+1)+(n+1)\right)\right)=$$

$$=\sin\left(\pi\cdot\left(\lim_{n\rightarrow\infty}\left(\sqrt{n^2+n+1}-(n+1)\right)+\lim_{n\rightarrow\infty}\left(n+1\right)\right)\right)=$$

$$=\sin\left(\pi\cdot\lim_{n\rightarrow\infty}\left(-\frac12+(n+1)\right)\right)=$$

$$=\lim_{n\rightarrow\infty}\sin\left(\pi n+\frac{\pi}{2}\right)=\lim_{n\rightarrow\infty}(-1)^{n}$$

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    $\begingroup$ There are several major mistakes in the above, imo: first, to use arithmetic of limits one must have that each of the expression part's limit exists finitely. This is not the case with $\;\lim (n+1)\;$ . Second, the one before last line has the expression $\;n+1\;$ there, so what happened with the limit?? Third, in order to exchange limit and function as done in the first line, one must either have the function is continuous where $\;n\;$ tends or prove otherwise the change is possible. None is done/fulfilled here. $\endgroup$ – Timbuc Jan 1 '15 at 17:41
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    $\begingroup$ You also need to prove that $$\lim_{n\to\infty}\bigg(\sqrt{n^2+n+1}-(n+1)\bigg) = -\frac{1}{2}$$ $\endgroup$ – John Joy Jan 1 '15 at 17:42
  • $\begingroup$ The very tiny correction made to this answer doesn't address the above questions. $\endgroup$ – Timbuc Jan 1 '15 at 17:47

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