text: # 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
  $\displaystyle\sum\limits_{n=1}^{\infty} \frac{\alpha^n}{\sqrt{n}}$
  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.
  • Aug 18th 2011, 07:14 AM
  Jester
  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.
  • 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
  
  $\lim_{n \to \infty} |\alpha \frac{\sqrt{n}}{\sqrt{n+1}}| < 1$
  
  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
  • 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 $n=0$, 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.
  • Aug 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?
  • Aug 18th 2011, 10:55 AM
  Jester
  Re: Prove this series converges
  Nope - it doesn't matter.