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 December 29th, 2010, 10:43 PM #1 Member   Joined: Oct 2010 Posts: 37 Thanks: 0 convergence of series `For what values of the real number $a$ does the following series converge? $\sum_{n=1}^{\infty} (\frac{1}{n}-\sin\frac{1}{n})^a$
 December 30th, 2010, 06:28 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: convergence of series Hint 1: use the series expansion of sin. Hint 2: the harmonic series diverges.
 December 30th, 2010, 08:03 AM #3 Member   Joined: Oct 2010 Posts: 37 Thanks: 0 Re: convergence of series I get $(\frac{\frac{1}{n^3}}{3!} - \frac{\frac{1}{n^5}}{5!} + \frac{\frac{1}{n^7}}{7!} - ....)^a$. Then? And it is not a harmonic series. (I think a harmonic series is given by $\sum_{n=1}^{\infty}\frac{1}{n}$)
December 30th, 2010, 08:15 AM   #4
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Re: convergence of series

Quote:
 Originally Posted by Sambit I get $(\frac{\frac{1}{n^3}}{3!} - \frac{\frac{1}{n^5}}{5!} + \frac{\frac{1}{n^7}}{7!} - ....)^a$. Then? And it is not a harmonic series. (I think a harmonic series is given by $\sum_{n=1}^{\infty}\frac{1}{n}$)
So you have something that looks like $\sum_{n=1}^\infty\left(\frac{1}{6n^3}\right)^a+\te xt{small stuff}.$ If a = 1, this obviously converges to something like zeta(3)/6 + small stuff. What would a have to be to make this diverge?

 December 30th, 2010, 08:25 AM #5 Member   Joined: Oct 2010 Posts: 37 Thanks: 0 Re: convergence of series greater than 3
December 30th, 2010, 09:35 AM   #6
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Re: convergence of series

Quote:
 Originally Posted by Sambit greater than 3
Well, the larger a is, the smaller all the terms would be.

 December 30th, 2010, 09:35 AM #7 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: convergence of series You're thinking along the right lines, but that would converge to zeta(9)/k + small stuff. Edit: DLowry beat me to it.
 December 30th, 2010, 09:52 PM #8 Member   Joined: Oct 2010 Posts: 37 Thanks: 0 Re: convergence of series silly mistake on my part. it diverges when $a\leq\frac{1}{3}$
 December 30th, 2010, 10:00 PM #9 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: convergence of series Ding ding ding, give that man a prize!
 January 2nd, 2011, 01:08 PM #10 Member   Joined: Nov 2010 Posts: 77 Thanks: 0 Re: convergence of series Hi the sum is < <$-\frac{\text{Zeta}[a]}{-1-a}+\frac{a \text{Zeta}[a]}{-1-a}$ then when a<=1 the series diverge

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