# Test for convergence/divergence of $\sum_{n=1}^{\infty}\frac{\sqrt{n}\cos(n)}{n+8}$

Given the series

$$\sum_{n=1}^{\infty}\frac{\sqrt{n}\cos(n)}{n+8}$$

I just came across the following question from an book. I need to test for convergence/divergence. My guess it is convergent, but how to show it?

Any help is appreciated. Thanks.

• What are some tests you know about? – user61527 Apr 26 '14 at 21:47
• Dirichlet's Test will be helpful here (en.wikipedia.org/wiki/Dirichlet's_test) – Davis Yoshida Apr 26 '14 at 21:48
• @DavisYoshida: What is the uniform (with respect to $N$) bound of all the $N^\text{th}$ partial sums, $\sum_{n=1}^N \cos n$? – Eric Towers Apr 26 '14 at 21:53
• @EricTowers $\sum\limits_{n=1}^{N}\cos{n} = \frac{\sin{\left(\frac{N+1}{2}\right)} - \sin{\left(\frac{1}{2}\right)}}{2\sin{\left(\frac{1}{2}\right)}}$ (which is clearly bounded) – Davis Yoshida Apr 26 '14 at 22:01

Let $a_n = \dfrac{\sqrt{n}}{n+8}$, $b_n = \cos n$. Clearly $\exists n_0 \in \mathbb{N}$ such that $a_n$ decreases monotonically $\forall n \geq n_0$, and $\displaystyle{\lim _{n \rightarrow \infty} a_n } = 0$.
Moreover, $\exists M \geq 0 : $$\left |\displaystyle{\sum _{n=n_0} ^ N \cos n} \right |\leq M \forall N \geq n_0. Therefore, using Dirichlet's test \displaystyle {\sum _{n=n_0} ^\infty a_n b_n} converges, and therefore \displaystyle {\sum _{n=1} ^\infty a_n b_n} converges too. NOTE: To prove that C_N = \displaystyle{ \sum_{n=0} ^{N} \cos(n)} is bounded, let's consider S_N = \displaystyle{ \sum_{n=0} ^{N} e^{in}}. Then, C_N = \Re (S_N), so we just have to prove that |S_N| is bounded.$$S_n = \displaystyle{ \sum_{n=0} ^{N} e^{in}} = \dfrac{1-e^{iN}}{1-e^i} |S_N | = \left | \dfrac{1-e^{iN}}{1-e^i} \right | = \dfrac{|1-e^{iN}|}{|1-e^i|}$$Now, |z+w|^2 \leq (|z| + |w|)^2 \forall z, w \in \mathbb{C}, so:$$|S_N| \leq \dfrac{(1+1)^2}{|1-e^i|} = \dfrac{4}{|1-e^i|}$$So$|S_N|\$ is bounded and we're done.