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Help me please to prove absolute and conditional convergence of: $$ \sum_{n=2}^{\infty }\frac{\sin (n+\frac{\pi }{3})}{\ln(n)} $$

Thanks a lot!

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Do you know Abel's transform? –  Davide Giraudo Feb 7 '12 at 9:32
    
No, i don't know –  Lilly Feb 7 '12 at 11:09

1 Answer 1

up vote 1 down vote accepted

Some steps, since it's a homework problem.

Convergence of the series

  1. Put $s_n:=\sum_{k=0}^n\sin(k+\frac{\pi}3)$. Show that we can find a constant $M>0$ such that $|s_n|\leq M$ for all $n\in\mathbb N$.
  2. Writing $$\sum_{n=1}^N\frac{\sin(n+\frac{\pi}3)}{\ln n}=\sum_{n=1}^N\frac{s_n-s_{n-1}}{\ln n}=\sum_{j=1}^N\frac{s_j}{\ln j}-\sum_{j=0}^{N-1}\frac{s_j}{\ln(j+1)},$$ show that the sequence $\left\{\sum_{n=1}^N\frac{\sin(n+\frac{\pi}3)}{\ln n}\right\}$ is Cauchy, hence convergent.

Absolute convergence of the series

  1. Using $\sin^2t\leq |\sin t|$, show that $$\frac{|\sin(n+\frac{\pi}3)|}{\ln n}\geq \frac 12\left(\frac 1{\ln n}-\frac{\cos(2(n+\frac{\pi}3))}{\ln n}\right)\geq 0.$$
  2. Show that the series $\sum_{n=1}^{+\infty}\frac{\cos(2(n+\frac{\pi}3))}{\ln n}$ is convergent.
  3. Conclude.
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@:It's not hw problem at all, thanks, but can i prove it using more simple way? Your answer about absolute convergence seems to me too complicated ... –  Lilly Feb 7 '12 at 14:12
    
Maybe there is a more simple way, but I didn't find it. Maybe an other user will manage to do this. In any case, the part about absolute convergence is just a little more complicated than the convergence, since question 2 can be solved in the same way than in the first part, and question 1 is just manipulating trigonometrical identities. –  Davide Giraudo Feb 7 '12 at 15:43
    
@ Davide Giraudo So I have that is no absolute convergence but it is convergent conditionally, right? –  Lilly Feb 7 '12 at 19:36
    
Yes, that's right. –  Davide Giraudo Feb 7 '12 at 19:45
    
How does showing that \frac{|\sin(n+\frac{\pi}3)|}{\ln n} is greater than something that converges absolutely show that it converges absolutely? –  Angela Richardson Feb 8 '12 at 13:21

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