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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!

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    All four series are _absolutely_ convergent. – 2017-02-01
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    @DanielFischer Well, crap... could you also provide some step solutions? – 2017-02-01
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    But (3) doesn't response to ratio cri. – 2017-02-01
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    @MyGlasses Then what criterion would you suggest? Dirichlet? – 2017-02-01
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    This series is gamma euler. – 2017-02-01

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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.

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    I had no idea about the relation between sin and abs. Thank alot, mate! – 2017-02-01
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    Quick question: For a) and b), when contructing the comparison series, did you just elude the $(-1)^n$ – 2017-02-01
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    One could say so. Since I argue for absolute convergence, I (tacitly) look at the absolute value of the terms. For the first two series that amounts to dropping the $(-1)^n$ resp. $(-1)^{n+1}$, for the last two, it amounts to nothing since the terms are positive anyway. – 2017-02-01