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