convergence or Divergence of Alternate Series

Question

Finding whether the series $$\sum^{\infty}_{n=1}\frac{(-1)^n}{4^n\cdot n^{\frac{3}{2}}}$$

What i try: $$a_{n}=\frac{(-1)^n}{4^n\cdot n^{\frac{3}{2}}}$$

And $$a_{n+1}=\frac{(-1)^{n+1}}{4^{n+1}\cdot (n+1)^{\frac{3}{2}}}$$

Using ratio test

$$\lim_{n\rightarrow \infty}(-1)^n\cdot \frac{1}{4}\bigg(\frac{n}{n+1}\bigg)^{\frac{3}{2}}$$

Now How do i solve that problem.

i did not understand How do i solve it.

Help me please

Answer 1

Now, you use the fact that$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac14\left(\frac n{n+1}\right)^{3/2}=\frac14<1$$and that therefore your series converges absolutely.

Answer 2

The Leibniz test tells us that the series converges, because it's alternating and decreasing.