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

$$\sum_{n\ge1} \sin (\pi \sqrt{n^2+1}) $$

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


I'm going to post the solution I've found here:

$$a_{n}= \sin (\pi \sqrt{n^2+1})=\sin (\pi (\sqrt{n^2+1}-n)+n\pi)=(-1)^n \sin (\pi (\sqrt{n^2+1}-n))=$$ $$ (-1)^n \sin \frac{\pi}{\sqrt{n^2+1}+n}$$ The sequence $b_{n} = \sin \frac{\pi}{\sqrt{n^2+1}+n}$ monotonically decreases to $0$. Since our series is an alternating series then it converges.

  • $\begingroup$ more than what? $\endgroup$
    – joriki
    Jul 14, 2012 at 16:44

2 Answers 2


$$ \sum_{n\ge1} \sin\left(\pi\sqrt{n^2+1}\right) = \sum_{n\ge1} \pm\sin\left(\pi\left(\sqrt{n^2+1}-n\right)\right) $$ (Trigonometric identity. Later we'll worry about "$\pm$".)

Now $$ \sqrt{n^2+1}-n = \frac{1}{\sqrt{n^2+1}+n} $$ by rationalizing the numerator.

So we have the sum of terms whose absolute values are $$ \left|\sin\left(\frac{\pi}{\sqrt{n^2+1}+n}\right) \right| \le \frac{\pi}{\sqrt{n^2+1}+n} \to0\text{ as }n\to\infty.\tag{1} $$

But the signs alternate and the terms decrease in size, so this converges. (They decrease in size because sine is an increasing function near $0$ and the sequence inside the sine decreases.)

It does not converge absolutely, since $\sin x\ge x/2$ for $x$ small and positive, and the sum of the terms asserted to approach $0$ in $(1)$ above diverges to $\infty$.

  • 4
    $\begingroup$ You need to show that the absolute value of the terms are decreasing too, not just that they tend to 0. $\endgroup$
    – mrf
    Jul 14, 2012 at 17:09
  • $\begingroup$ @mrf : Maybe you're right ---- I wonder if one could get by with something weaker than "decreasing". But I think we can show this is decreasing. $\endgroup$ Jul 14, 2012 at 17:13
  • $\begingroup$ @mrf : OK, I've now edited my answer to mention that they are readily seen to decrease. Thanks for pointing this out. $\endgroup$ Jul 14, 2012 at 17:15
  • 1
    $\begingroup$ Much better :) In this case it wasn't hard to see that the terms are decreasing, but it's such a common mistake by students not to check the monotonicity requirement in Leibniz' criterion that I wanted to point it out. $\endgroup$
    – mrf
    Jul 14, 2012 at 17:16

It converges as an alternating series. For each $n$, we have $\sin((n+\delta)\cdot\pi)$ for some small $\delta$ which approaches $0$ at the limit and decreases monotonically (as the $1$ in $\sqrt{n^2+1}$ becomes less significant compared to the $n^2$.

For even $n$, this expression will take on smaller and smaller positive values, as $\delta$ shrinks and $(n+\delta)\cdot\pi$ gets closer and closer to a zero at $2m\pi$ for some natural $m$ from the right.

Similarly, it will take on shrinking negative values for odd $n$.

The absolute value of each term tends to $0$ and decreases monotonically, so we have convergence by the alternating series test.

  • $\begingroup$ Nice response!! $\endgroup$ Jul 27, 2012 at 2:44

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