Hi,

I have the following question:

Part d

Using appropriate series convergence tests, prove that the series

$\displaystyle $\displaystyle\sum\limits_{n=1}^{\infty} \frac{\alpha^n}{\sqrt{n}}$$

converges if and only if the real number $\displaystyle \alpha$ satisfies

$\displaystyle -1\le\alpha<1$

So my current working is this:

Let $\displaystyle a_n=\frac{\alpha^n}{\sqrt{n}}$, then

$\displaystyle r= $$\displaystyle \frac{a_{n+1}}{a_n}$$\displaystyle =\frac{\alpha\sqrt{n}} {\sqrt{n+1}}$

Now as $\displaystyle n$ tends to infinity, $\displaystyle r$ will tend to $\displaystyle \alpha$.

Therefore, by the ratio test:

If $\displaystyle \alpha<1$, the series converges

If $\displaystyle \alpha>1$, the series diverges

If $\displaystyle \alpha=1$, $\displaystyle a_n$$\displaystyle =\frac{1}{\sqrt{n}}$

In part b of this same question, we had to prove that $\displaystyle \frac{1}{\sqrt{n}}$ diverges, which I did successfully.

According to my working then, the series converges if $\displaystyle \alpha<1$, which only partly satisfies the requirement: $\displaystyle -1\le\alpha<1$

Can someone spot what I've done wrong here?

Thanks

Note: In part c of the same question, we had to state and prove the Alternating Series Test, which I did successfully, so I'm thinking that we're meant to apply it here somehow, but I'm not sure how.