# Prove this series converges

• Aug 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
$\displaystyle$\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$Can someone spot what I've done wrong here? Thanks Note: In part c of the same question, we had to state and prove the Alternating Series Test, which I did successfully, so I'm thinking that we're meant to apply it here somehow, but I'm not sure how. • Aug 18th 2011, 07:14 AM Jester Re: Prove this series converges Actually what you want using the ratio test is$\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}}$Then determine the convergence of each one to determine whether these endpoints should be included in your interval. • Aug 18th 2011, 10:41 AM dwally89 Re: Prove this series converges Quote: Originally Posted by Danny Actually what you want using the ratio test is$\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$Is this the same as what Wikipedia says? Thanks again • Aug 18th 2011, 10:46 AM Jester Re: Prove this series converges Quote: Originally Posted by dwally89 Thanks. According to Alternating series test - Wikipedia, the free encyclopedia, the Alternating Series Test is from$\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.$As for the test, it doesn't matter shere you start. Adding or subtracting a few terms from the series won't change whether the series converges or not. • Aug 18th 2011, 10:51 AM dwally89 Re: Prove this series converges Quote: Originally Posted by Danny You'll notice that$\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?
• Aug 18th 2011, 10:55 AM
Jester
Re: Prove this series converges
Nope - it doesn't matter.