Prove this series converges

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.

Re: Prove this series converges

Actually what you want using the ratio test is

$\displaystyle \lim_{n \to \infty} |\alpha \frac{\sqrt{n}}{\sqrt{n+1}}| < 1 $

which gives $\displaystyle | \alpha | < 1$ or $\displaystyle - 1 < \alpha < 1$. Then check your endpoints giving the two series

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$

and

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$

Then determine the convergence of each one to determine whether these endpoints should be included in your interval.

Re: Prove this series converges

Quote:

Originally Posted by

**Danny** Actually what you want using the ratio test is

$\displaystyle \lim_{n \to \infty} |\alpha \frac{\sqrt{n}}{\sqrt{n+1}}| < 1 $

which gives $\displaystyle | \alpha | < 1$ or $\displaystyle - 1 < \alpha < 1$. Then check your endpoints giving the two series

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$

and

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$

Then determine the convergence of each one to determine whether these endpoints should be included in your interval.

Thanks.

According to Alternating series test - Wikipedia, the free encyclopedia, the Alternating Series Test is from $\displaystyle n=0$, not $\displaystyle n=1$. Does this make a difference?

I previously knew the Alternating Series Test as

$\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$

Is this the same as what Wikipedia says?

Thanks again

Re: Prove this series converges

Quote:

Originally Posted by

**dwally89** Thanks.

According to

Alternating series test - Wikipedia, the free encyclopedia, the Alternating Series Test is from $\displaystyle n=0$, not $\displaystyle n=1$. Does this make a difference?

I previously knew the Alternating Series Test as

$\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$

Is this the same as what Wikipedia says?

Thanks again

You'll notice that $\displaystyle \frac{1}{\sqrt{n}} $ is not defined at $\displaystyle n = 0$ so it should start at $\displaystyle n = 1.$ As for the test, it doesn't matter shere you start. Adding or subtracting a few terms from the series won't change whether the series converges or not.

Re: Prove this series converges

Quote:

Originally Posted by

**Danny** You'll notice that $\displaystyle \frac{1}{\sqrt{n}} $ is not defined at $\displaystyle n = 0$ so it should start at $\displaystyle n = 1.$ As for the test, it doesn't matter shere you start. Adding or subtracting a few terms from the series won't change whether the series converges or not.

Is "my version" of the alternating series test:

$\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$

wrong then?

Re: Prove this series converges

Nope - it doesn't matter.