yet another series question: sum[n=1,infty][sin(1/n^2)ln(n)]

[cheffy]

Last one, I swear!

\(\displaystyle \
\sum\limits_{n = 1}^\infty {\sin \left( {\frac{1}{{n^2 }}} \right)\ln (n)}
\\)

Does this converge or diverge?

I can't substitute so what test would I do?

Thank you!

 

[galactus]

I tried the integral test.

\(\displaystyle \L\\\int_{1}^{\infty}[sin(\frac{1}{n^{2}})ln(n)]dn=0.993435049565\)

Since the integral converges, so doth the series.

I ran this through Maple. The definite integral to this was awesome. It was huge and had terms in it I am not familiar with. If I can, I will post it for kicks.