Does the following series converge?
$$\sum_{n=1}^{\infty}\frac{\cos\left({\frac{n\pi}{2}}\right)}{\sqrt{n}}$$
The $\cos$ function:
- alternates between (-1) and 1 for every $n$ that is even. (for a general expression see this)
- equal to zero for every $n$ that is odd.
Thus the sequence of the given series is actually alternating between positive and negative:
$$
\sum_{n=1}^{\infty}\frac{\cos\left({\frac{n\pi}{2}}\right)}{\sqrt{n}} = 0 + \frac{-1}{\sqrt{2}} + 0 + \frac{1}{\sqrt{4}} + 0 + \frac{-1}{\sqrt{6}} + \cdots + \frac{i^{n} (1 + (-1)^{n})}{2}\cdot\frac{1}{\sqrt{n}}
$$
Would it be valid to use the Leibniz Test in order to say that the given series converge?
Do you know of a different method to prove whether the given series converges/diverges? Without using the complex number $i$.
Best Answer
In the Leibniz Test, for the convergence of $\sum(-1)^n a_n$ is required $a_n\ge 0$, $a_n\to 0$ and $a_{n+1}\le a_n$, so take only the even terms. And the sum isn't zero.