Prove this series converges

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August 18th 2011, 05:42 AM
dwally89

Prove this series converges

Hi,

I have the following question:

Part d
Using appropriate series convergence tests, prove that the series
$\displaystyle\sum\limits_{n=1}^{\infty} \frac{\alpha^n}{\sqrt{n}}$
converges if and only if the real number $\alpha$ satisfies
$-1\le\alpha<1$

So my current working is this:

Let $a_n=\frac{\alpha^n}{\sqrt{n}}$ , then
$r=$ $\frac{a_{n+1}}{a_n}$ $=\frac{\alpha\sqrt{n}} {\sqrt{n+1}}$

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

Therefore, by the ratio test:
If $\alpha<1$ , the series converges
If $\alpha>1$ , the series diverges
If $\alpha=1$ , $a_n$ $=\frac{1}{\sqrt{n}}$
In part b of this same question, we had to prove that $\frac{1}{\sqrt{n}}$ diverges, which I did successfully.

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

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.
August 18th 2011, 07:14 AM
Danny

Re: Prove this series converges

Actually what you want using the ratio test is

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

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

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

and

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

Then determine the convergence of each one to determine whether these endpoints should be included in your interval.
August 18th 2011, 10:41 AM
dwally89

Re: Prove this series converges

Quote:

Originally Posted by Danny

Actually what you want using the ratio test is

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

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

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

and

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

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 $n=0$ , not $n=1$ . Does this make a difference?

I previously knew the Alternating Series Test as
$\sum_{n=1}^{\infty}(-1)^{n+1}a_n$
Is this the same as what Wikipedia says?

Thanks again
August 18th 2011, 10:46 AM
Danny

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 $n=0$ , not $n=1$ . Does this make a difference?

I previously knew the Alternating Series Test as
$\sum_{n=1}^{\infty}(-1)^{n+1}a_n$
Is this the same as what Wikipedia says?

Thanks again

You'll notice that $\frac{1}{\sqrt{n}}$ is not defined at $n = 0$ so it should start at $n = 1.$ 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.
August 18th 2011, 10:51 AM
dwally89

Re: Prove this series converges

Quote:

Originally Posted by Danny

You'll notice that $\frac{1}{\sqrt{n}}$ is not defined at $n = 0$ so it should start at $n = 1.$ 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:
$\sum_{n=1}^{\infty}(-1)^{n+1}a_n$
wrong then?
August 18th 2011, 10:55 AM
Danny

Re: Prove this series converges

Nope - it doesn't matter.