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azatkgz
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[SOLVED] Series with Trigonometric funtions
Determine whether the series converges conditionally,converges absolutely or diverges.
\sum_{n=2}^{\infty}\frac{\sin (n+\frac{1}{n})}{\ln (\ln n)}
\frac{\sin (n+\frac{1}{n})}{\ln (\ln n)}=\frac{\sin n\cos \frac{1}{n}+\sin \frac{1}{n}\cos n}{\ln (\ln n)}=\frac{\sin n(1+O(\frac{1}{n^2}))+\cos n(\frac{1}{n}+O(\frac{1}{n^3}))}{\ln (\ln n)}
\sum_{n=2}^{\infty}\sin n,\sum_{n=2}^{\infty}\cos n are bounded
From Dirichlet's Test we can deduce that this series converges
|a_n|=\left|\frac{\sin n(1+O(\frac{1}{n^2})}{\ln (\ln n)}\right|+\left|\frac{cos n(\frac{1}{n}+O(\frac{1}{n^3}))}{\ln (\ln n)}\right|
if we look for example at \left|\sum_{n=2}^{\infty}\frac{cos n}{n\ln (\ln n)}\right|>\left|\sum_{n=2}^{\infty}\frac{cos n}{n\ln (n)}\right|
0<|cosn|<1 and \sum_{n=2}^{\infty}\frac{1}{n\ln (n)} diverges by Integral Test
So this series converges conditionally!
Homework Statement
Determine whether the series converges conditionally,converges absolutely or diverges.
\sum_{n=2}^{\infty}\frac{\sin (n+\frac{1}{n})}{\ln (\ln n)}
The Attempt at a Solution
\frac{\sin (n+\frac{1}{n})}{\ln (\ln n)}=\frac{\sin n\cos \frac{1}{n}+\sin \frac{1}{n}\cos n}{\ln (\ln n)}=\frac{\sin n(1+O(\frac{1}{n^2}))+\cos n(\frac{1}{n}+O(\frac{1}{n^3}))}{\ln (\ln n)}
\sum_{n=2}^{\infty}\sin n,\sum_{n=2}^{\infty}\cos n are bounded
From Dirichlet's Test we can deduce that this series converges
|a_n|=\left|\frac{\sin n(1+O(\frac{1}{n^2})}{\ln (\ln n)}\right|+\left|\frac{cos n(\frac{1}{n}+O(\frac{1}{n^3}))}{\ln (\ln n)}\right|
if we look for example at \left|\sum_{n=2}^{\infty}\frac{cos n}{n\ln (\ln n)}\right|>\left|\sum_{n=2}^{\infty}\frac{cos n}{n\ln (n)}\right|
0<|cosn|<1 and \sum_{n=2}^{\infty}\frac{1}{n\ln (n)} diverges by Integral Test
So this series converges conditionally!
