<s>[INST] limit and summation I have to calculate $$ \lim_{m\rightarrow\infty}\sum_{n=0}^{\infty}\frac{1}{m^n} \sum_{k=0}^{\infty}  {2k \choose k }  {2n \choose k}\left( \frac{-1}{2} \right)^k$$
We know that $$ \sum_{k=0}^{2n}  {2k \choose k }  {2n \choose k}\left( \frac{-1}{2} \right)^k=\frac{1}{4^n}\binom{2n}{n}=\frac1\pi\int_0^1\frac{x^n\,\mathrm{d}x}{\sqrt{x(1-x)}}$$
so
$$ \sum_{k=0}^{\infty}  {2k \choose k }  {2n \choose k}\left( \frac{-1}{2} \right)^k=\lim_{n\rightarrow\infty} \sum_{k=0}^{2n}  {2k \choose k }  {2n \choose k}\left( \frac{-1}{2} \right)^k=\lim_{n\rightarrow\infty}\frac{1}{4^n}\binom{2n}{n}=0$$
and
$$ \sum_{n=0}^{\infty}\frac{1}{m^n} \sum_{k=0}^{\infty}  {2k \choose k }  {2n \choose k}\left( \frac{-1}{2} \right)^k = \sum_{n=0}^{\infty}0=0$$
Am I right?
 [/INST] For third limit, you could also have used that, by the Wallis product,
$$\binom{2 n}{n} \sim \frac {4^n} {\sqrt{\pi n}}$$ making
$$\frac{1}{4^n}\binom{2n}{n}\sim \frac {1} {\sqrt{\pi n}}$$ 
Remember also that there exist tight bounds for the central binomial coefficient
$$\frac {4^n} {\sqrt{4 n}} \leq \binom{2 n}{n} \leq\frac {4^n} {\sqrt{3 n+1}}\qquad \forall n \geq 1 $$ and, even more accurate
$$\binom{2 n}{n}\sim \frac {4^n} {\sqrt{\pi n}} \left(1-\frac {c_n} n \right)\qquad \text{where} \qquad \frac 19 < c_n <\frac 18 \qquad \forall n \geq 1$$
Using the very first approximation,we then have
$$\sum_{n=0}^{\infty}\frac{1}{m^n\sqrt{\pi n}}=\frac{1}{\sqrt{\pi }}\text{Li}_{\frac{1}{2}}\left(\frac{1}{m}\right)=\frac{1}{\sqrt{\pi }}\sum_{k=1}^\infty \frac 1 {\sqrt{k} \,m^k}$$
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