I have a problem with determining the absolute convergence:
$$ \sum_{n=1}^\infty \left(\cos\left(\ln\left(\frac{n+1}{n}\right)\right) - n\sin\left(\frac{1}{n}\right)\right) $$
I assume that $n\sin(\frac{1}{n})\to 1$, likewise $\cos(\ln(\frac{n+1}{n}))$ because $\ \ln(\frac{n+1}{n}) = \ln(1+\frac{1}{n})\to 0$, thus we have $\ \cos(0) = 1$
However, we get $1-1$. How can I calculate which is quicker in approaching?
EDIT: I have an idea that I have to compare with $\frac{1}{n^2}$, which is convergent. If $\frac{\displaystyle\sum_{n=1}^\infty\left(\cos\left(\ln\left(\frac{n+1}{n}\right)\right) - n\sin\left(\frac{1}{n}\right)\right)}{\frac{1}{n^2}} $ would have posotove finit limit, it would converge, however I am stuck on calculation because I don't know how to solve $\ n^2\left(\cos\left(\ln\left(1+\frac{1}{n}\right)\right) - n\sin\left(\frac{1}{n}\right)\right) $
You need to approximate everything to the second order:
$\sin x\approx x-\frac16x^3$, so $n\sin\frac1n\approx 1-\frac{1}{6n^2}$.
$\ln(1+x)\approx x-\frac12x^2$, so $\ln(\frac{n+1}{n})\approx \frac1n-\frac{1}{2n^2}$.
And $\cos x\approx 1-\frac12x^2$, so $\cos(\ln(\frac{n+1}{n}))\approx 1-\frac12(\frac1n-\frac{1}{2n^2})^2\approx 1-\frac{1}{2n^2}$.
So you get $\cos(\ln(\frac{n+1}{n}))-n\sin\frac1n\approx -\frac{1}{3n^2}$. And $\sum\frac{1}{3n^2}$ converges absolutely.
Turning this into a rigorous proof is a bit messy, but not very complicated $-$ you just have to keep track of the error terms, which are all $O(\frac{1}{n^3})$ or less.