determine series convergence.

Question

Determine whether this series converges:

$\sum_{n=1}^\infty \cos(n^2\pi) (\sqrt{n+11} -\sqrt{n+2}) $

I know that $\lim a_{n} = 0$ and that this series alternates because of $cos(n^2\pi)$, but don't know where to go from here.

Answer 1

Observe that $$|a_n|=\sqrt{n+11}-\sqrt{n+2}=\frac{n+11-n-2}{\sqrt{n+11}+\sqrt{n+2}}=\frac{9}{\sqrt{n+11}+\sqrt{n+2}}$$ and that this sequence is decreasing and converges to zero. So, Leibniz criterion for alternate series says that it is convergent.

Answer 2

Try to use the alternating series test. Note that $a_{n} = \sqrt{n+11} - \sqrt{n+2}$ is a decreasing sequence (why?) and converges to 0. Also, we have $\cos(n^{2}\pi) = (-1)^{n}$.

Answer 3

Since the square of an even number is an even number and the square of an odd number is an odd number, then $\cos(n^2\pi)=(-1)^n$. Now appeal to Leibniz's Test.

Can you finish?