Another limit related to pi number Find the value of the limit:
$$\lim_{n\to\infty} \sum_{k=0}^n \frac{{k!}^{2} {2}^{k}}{(2k+1)!}$$
I'm trying to find out if this limit can be computed only by using high school 
knowledge for solving limits. Thanks. 
 A: Mimicking robjohn's solution to the series, and after proving convergence, we may proceed as follows:
$$\sum\limits_{k = 0}^\infty  {\frac{{k!^2{2^k}}}{{\left( {2k + 1} \right)!}}}  = \sum\limits_{k = 1}^\infty  {\frac{{\left( {k - 1} \right)!^2{2^{k - 1}}}}{{\left( {2k - 1} \right)!}}}  = \sum\limits_{k = 1}^\infty  {\frac{{{\Gamma ^2}\left( k \right)}}{{\Gamma \left( {2k} \right)}}{2^{k - 1}}}  = \sum\limits_{k = 1}^\infty  {\operatorname{B} \left( {k,k} \right){2^{k - 1}}} $$ 
$$\sum\limits_{k = 1}^\infty  {\operatorname{B} \left( {k,k} \right){2^{k - 1}}}  = \sum\limits_{k = 1}^\infty  {\int\limits_0^1 {{{\left[ {2t\left( {1 - t} \right)} \right]}^{k - 1}}dt} }  = \int\limits_0^1 {\sum\limits_{k = 1}^\infty  {{{\left[ {2t\left( {1 - t} \right)} \right]}^{k - 1}}} dt} $$
Then
$$=\int\limits_0^1 {\frac{{dt}}{{1 - 2t\left( {1 - t} \right)}}}  = \int\limits_0^1 {\frac{{dt}}{{1 - 2t + 2{t^2}}}}  = \frac{1}{2}\int\limits_0^1 {\frac{{dt}}{{{{\left( {t - \frac{1}{2}} \right)}^2} + \frac{1}{4}}}} $$
Now let $t-1/2=u$.
$$\frac{1}{2}\int\limits_{ - 1/2}^{1/2} {\frac{{du}}{{{u^2} + {{\left( {1/2} \right)}^2}}}}  = \arctan 2\frac{1}{2} - \arctan 2\left( { - \frac{1}{2}} \right)$$
$$=2\arctan 1=2\frac{\pi}{4}=\frac{\pi}{2}$$
A: This is a formula of Euler (1737) giving $\frac {\pi}2$. A solution and a proof using the expansion of arctan may be found in Boris Gourévitch's 'World of pi'.
The following discussion could help too
A: If students don't know about the Gauss error function -- which is defined in terms of a non-elementary integral -- then no! Because the exact value of this infinite sum is $$\sqrt\frac{e \pi}2  \operatorname{erf}\left(\frac{1}{\sqrt 2}\right)$$.
