Originally Posted by

**galactus** The first one converges.

If you are curious:

Look at the relation: $\displaystyle \frac{a_{2n}}{a_{n}}=\frac{a_{n+1}}{a_{n}}\cdot\fr ac{a_{n+2}}{a_{n+1}}\cdot\cdot\cdot\frac{a_{2n}}{a _{2n-1}}$

$\displaystyle \frac{a_{2n+1}}{a_{n}}=\frac{a_{n+1}}{a_{n}}\cdot\ frac{a_{n+2}}{a_{n+1}}\cdot\cdot\cdot\frac{a_{2n+1 }}{a_{2n}}$

Then $\displaystyle \frac{a_{2n+1}}{a_{n}}<\frac{a_{2n}}{a_{n}}$

$\displaystyle =\frac{(2n+1)(2n+3)\cdot\cdot\cdot(4n-1)}{2^{n}(n+2)(n+3)\cdot\cdot\cdot(2n)(2n+1)}$

$\displaystyle =\frac{(2n+3)(2n+5)\cdot\cdot\cdot(4n-1)}{2^{n}(n+2)(n+3)\cdot\cdot\cdot(2n)}$

$\displaystyle =\frac{1}{2}\left(\frac{2n+3}{2n+4}\right)\left(\f rac{2n+5}{2n+6}\right)\cdot\cdot\cdot\left(\frac{4 n-1}{4n}\right)$

$\displaystyle <\frac{1}{2}\left(\frac{4n-1}{4n}\right)^{n-1}=\frac{1}{2}\left(1-\frac{1}{4n}\right)^{n-1}$

Since $\displaystyle \lim_{n\rightarrow{\infty}}\left(1-\frac{1}{4n}\right)^{n-1}=\frac{1}{e^{\frac{1}{4}}}$

$\displaystyle \limsup_{n\rightarrow{\infty}}\frac{a_{2n+1}}{a_{n }}\leq\limsup_{n\rightarrow{\infty}}\frac{a_{2n}}{ a_{n}}\leq\frac{1}{2e^{\frac{1}{4}}}<\frac{1}{2}$

Therefore, the series converges.

The second one converges if p>2 and diverges if $\displaystyle p\leq{2}$

The third one, the harmonic, surprisingly converges.

It can be shown that if we delete from the harmonic series all terms that contain X, the resulting series in convergent.