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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

Martin Sleziak
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Educ
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3 Answers3

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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 :

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

Workaholic
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  • Why? Isn't $\sqrt{n^2+n+1} \approx \sqrt{n^2+2n+1} = (n+1)$ as $n \rightarrow \infty$ ? So $\sin(\pi\sqrt{n^2+n+1}) \rightarrow 0$ – Vladimir Fomenko Jan 01 '15 at 16:42
  • Why did you downvote? – Alex Silva Jan 01 '15 at 16:45
  • $$\lim_{n\to\infty}\bigg((n+1) - \sqrt{n^2+n+1}\bigg) = ?$$ – John Joy Jan 01 '15 at 16:51
  • @VolodymyrFomenko, you are wrong! This limit does not exist. See this in http://www.wolframalpha.com/. – Alex Silva Jan 01 '15 at 16:53
  • @VolodymyrFomenko $\sin$ is a periodic function. – Workaholic Jan 01 '15 at 16:54
  • @JohnJoy, good example. :) – Alex Silva Jan 01 '15 at 16:55
  • @user201168 i know that and i upvote only the detailed answers and i ask for different ways to prove that not just the hint – Educ Jan 01 '15 at 17:09
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    @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. – Timbuc Jan 01 '15 at 17:11
  • here is conterexample of what you said : user201168 is member for 19 days then he is newcomer and he just give the hint not detailed answer so what you say is generally not true – Educ Jan 01 '15 at 17:16
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    @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. – John Joy Jan 01 '15 at 17:29
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    @Educ, you can distinguish between "more usually" and "always", right? – Timbuc Jan 01 '15 at 17:36
  • @Timbuc, you can distinguish between "generally true" and "Always True ", right? – Educ Jan 01 '15 at 18:40
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    @Educ Perhaps I can, perhaps I cannot, yet what I wrote, and this is what you addressed, is "usually" . :) – Timbuc Jan 01 '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.

Aaron Maroja
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    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 $;\sin\pi(n+0.5);$ . –  Jan 01 '15 at 18:09
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$$\lim_{n\rightarrow\infty}{\sin\left(\pi\cdot\sqrt{n^2+n+1}\right)}=\sin{\lim_{n\rightarrow\infty}\left(\pi\cdot\sqrt{n^2+n+1}\right)}=$$

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

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

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

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

Vladimir Fomenko
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    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. – Timbuc Jan 01 '15 at 17:41
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    You also need to prove that $$\lim_{n\to\infty}\bigg(\sqrt{n^2+n+1}-(n+1)\bigg) = -\frac{1}{2}$$ – John Joy Jan 01 '15 at 17:42
  • The very tiny correction made to this answer doesn't address the above questions. – Timbuc Jan 01 '15 at 17:47