I'm trying some exercises on real number series, in which I have to see if the series are convergent or not:
$$a) \sum_{n=1}^\infty \ (-1)^n \frac{2n+1}{3^n}$$ $$b) \sum_{n=0}^\infty \ (-1)^{n+1} \frac{1}{(n+1)(2n)!}$$ $$c) \sum_{n=1}^\infty \ \frac{1}{n} - \ln(1+ \frac{1}{n})$$ $$d) \sum_{n=1}^\infty \sin \frac{1}{n(n+1)}$$
For the first three series I've found that:
a) is divergent (solved it using both the ration criterion and Dirichlet criterion as to verify the $1^{st}$ solution);
b) is divergent (solved same as $a)$);
c) is convergent (solved using the ration criterion).
For these three I'm not sure of my results and I hope that somebody here would point out if I'm mistaken regarding any of them. If that's the case, then I would kindly ask for an idea/ steps of a solution.
At last,
d) I've no idea how to solve it. I'd be very grateful if anyone could offer a solution for it, or just the main steps.
Thank you in advance!
All four of these series are absolutely convergent.
For $n \geqslant 3$ we have $2n+1 < 2^n$, so the term of the first series is majorised by $\bigl(\frac{2}{3}\bigr)^n$, and comparison with a geometric series shows absolute convergence.
We have $\frac{1}{(n+1)(2n)!} \leqslant \frac{1}{(n+1)!}$ since $n! \leqslant (2n)!$, and comparison with the exponential series shows absolute convergence.
For $x > 0$ we have
$$0 < x - \ln (1+x) < \frac{x^2}{2},$$
so comparison with $\sum \frac{1}{n^2}$ shows the convergence of the third series.
We have $\lvert \sin x\rvert \leqslant \lvert x\rvert$ for all $x\in \mathbb{R}$, so comparison with $\sum \frac{1}{n(n+1)}$ shows the convergence of the last series.