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Does the series converge? $$\sum_{n=1}^{\infty}(-1)^{n} \sin \left(\frac{\pi}{2}\left(1+\frac{1}{n}\right)\right)$$

I have tried to transform the series as follows

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

Using the fact $\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$ and then try to computate

$$\lim_{n \to \infty} a_{n}=\lim_{n \to \infty} (-1)^{n} \cos\left(\frac{\pi}{2n}\right)$$

But $(-1)^n$ doesn't exist. Can I be sure that this limit is not identically $0$ and therefore the series diverges?

Thanks for your help!

Martin Sleziak
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Isomorphicc
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    Well, $\sin \left( \dfrac{\pi}{2} \left( 1 + \dfrac{1}{n} \right) \right) \rightarrow 1$ so that $\left| \left( -1 \right)^n \sin \left( \dfrac{\pi}{2} \left( 1 + \dfrac{1}{n} \right) \right) \right| \not\rightarrow 0$ and hence the terms do not converge to $0$. – Aniruddha Deshmukh Oct 04 '21 at 04:09
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    @AniruddhaDeshmukh I'm sorry, can you explain this statement to me? I don't quite understand why you say that – Andre785858 Oct 04 '21 at 05:42
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    A sequence $\left( a_n \right)$ converges to $a$ if and only if $\left| a_n - a \right| \rightarrow 0$. In your case, the sequence $a_n = \left( -1 \right)^n \sin \left( \dfrac{\pi}{2} \left( 1 + \dfrac{1}{n} \right) \right)$ and $a = 0$. So, the sequence of terms converges to zero if and only if $\left| \left( -1 \right)^n \sin \left( \dfrac{\pi}{2} \left( 1 + \dfrac{1}{n} \right) \right) - 0 \right| = \left| \left( -1 \right)^n \sin \left( \dfrac{\pi}{2} \left( 1 + \dfrac{1}{n} \right) \right) \right| \rightarrow 0$, which is not the case. – Aniruddha Deshmukh Oct 04 '21 at 05:46
  • If you assume the sum converges after $M$ terms the next term will change the sum by approximately $+1$ or $-1$. The sum approximately oscillates. –  Oct 04 '21 at 08:07

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