# Convergence proof...

• Jun 27th 2010, 10:52 AM
Also sprach Zarathustra
Convergence proof...
Check the convergence of:

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

Thank you!
• Jul 1st 2010, 04:46 AM
CaptainBlack
Quote:

Originally Posted by Also sprach Zarathustra
Check the convergence of:

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

Thank you!

As $\displaystyle$$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:

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

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

So we need only consider the convergence of:

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

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

CB