# Is $\sum_{n=1}^{\infty} \frac{\sin(n) +\cos(n)}{n^2}$ convergent?

My apologies if this question is a duplicate. My attempt to show that it is convergent:

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

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

Since $$\sum_{n=1}^{\infty} \frac{1}{n^2}$$ is convergent then by the Comparison Test $$\sum_{n=1}^{\infty} \frac{\sin(n) + \cos(n)}{n^2}$$ is absolutely convergent. Therefore $$\sum_{n=1}^{\infty} \frac{\sin(n) + \cos(n)}{n^2}$$ is convergent.

I would like to know if the above is correct.

• I think your logic is correct, but could probably be written a bit more succinctly Commented Jul 30, 2021 at 21:22
• $\sin(n)+\cos(n) = \sqrt{2}\sin(n+\frac{\pi}{2})$ Commented Jul 30, 2021 at 22:58
• @haman_Abdallah: Thank you for this very useful comment; however, perhaps you meant $\sin(n+\frac{\pi}{4})$ instead of $\sin(n+\frac{\pi}{2})$ since $\sin(n+\frac{\pi}{4}) = \sin(n)/\sqrt 2 +\cos(n)/\sqrt 2$. Commented Jul 31, 2021 at 0:15

• You write that $$\displaystyle\sum_{n=1}^\infty\frac{\sin(n)+\cos(n)}{n^2}=\sum_{n=1}^\infty\frac{\sin(n)}{n^2}+\sum_{n=1}^\infty\frac{\cos(n)}{n^2}$$, but this equality assumes that the three series involved are convergent. However, you are supposed to prove that the first one converges.
• You write that $$\displaystyle\left|\sum_{n=1}^\infty\frac{\sin(n)+\cos(n)}{n^2}\right|\leqslant\left|\sum_{n=1}^\infty\frac{\sin(n)}{n^2}\right|+\left|\sum_{n=1}^\infty\frac{\cos(n)}{n^2}\right|$$, but, again, this assumes that you are dealing with numbers, that is, once again you are assuming that all those series are convergent.
You can simple say that$$(\forall n\in\Bbb N):\left|\frac{\sin(n)+\cos(n)}{n^2}\right|\leqslant\frac2{n^2}$$and then apply the comparison test. Note that this means to compare the terms of the given series with the terms of a convergent series.
• Thank you for the answer. I see my mistakes. Perhaps to arrive at the last inequality we should write that $(\forall n \in \mathbb{N})$: $\left| \frac{\sin(n) + \cos(n)}{n^2} \right| \leq \left| \frac{\sin(n)}{n^2} \right|+ \left| \frac{\cos(n)}{n^2} \right|$ to show where the 2'' in $\frac{2}{n^2}$ comes from. Commented Jul 31, 2021 at 0:24