Convergence proof...

**Also sprach Zarathustra** · June 27th 2010, 10:52 AM

Check the convergence of:

$\displaystyle \sum^{\infty}_{n=1} \int^{\frac{sin(n)}{n}}_0 \frac{sin(x)}{x}dx$

Thank you!

**CaptainBlack** · July 1st 2010, 04:46 AM

Originally Posted by Also sprach Zarathustra

Check the convergence of:

$\displaystyle \sum^{\infty}_{n=1} \int^{\frac{sin(n)}{n}}_0 \frac{sin(x)}{x}dx$

Thank you!

As $$$ n$ becomes large the intervals of integration contract about 0, where the integrand is close to 1. So we need only consider the convergence of:

$\displaystyle \sum^{\infty}_{n=1} \int^{\frac{sin(n)}{n}}_0 dx$

When $\sin(n)<0$ the corresponding term in the sum is $|\sin(n)|/n$ and when $\sin(n)>0$ the corresponding term is $\sin(n)/n$

So we need only consider the convergence of:

$\displaystyle \sum^{\infty}_{n=1}\frac{|sin(n)|}{n}$

which I'm pretty sure diverges, but will leave the proof of that to the reader

CB

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