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Evaluate Series
09-01-2010, 03:52 PM
Post: #1
Evaluate Series
(1) $\displaystyle{\sum_{n=1}^{\infty} \arctan \frac{1}{2k^2}}$
(2) $\displaystyle{\sum_{n=1}^{\infty} \frac{1}{2^{2n+1} n(2n-1)}}$
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10-08-2010, 04:06 PM
Post: #2
RE: Evaluate Series
(1) $\frac{1}{2k^2} =\displaystyle{ \frac{\frac{1}{k(k+1)}}{1+\frac{k-1}{k+1}}=\frac{\frac{k}{k+1}-\frac{k-1}{k}}{1+\frac{k}{k+1}\frac{k-1}{k}}}$ So $\arctan \frac{1}{2k^2} = \arctan{\frac{k}{k+1}}-\arctan{\frac{k-1}{k}}$
$\quad\quad$ Therefore $\displaystyle{\sum_{n=1}^{\infty} \arctan \frac{1}{2k^2} = \lim_{n\to\infty} \arctan\frac{k}{k+1}=1}$
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