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.
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 $t-1/2=u$.

$$\frac{1}{2}\int\limits_{ - 1/2}^{1/2} {\frac{{du}}{{{u^2} + {{\left( {1/2} \right)}^2}}}}  = \arctan 2\frac{1}{2} - \arctan 2\left( { - \frac{1}{2}} \right)$$

$$=2\arctan 1=2\frac{\pi}{4}=\frac{\pi}{2}$$
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 $$\sqrt\frac{e \pi}2  \operatorname{erf}\left(\frac{1}{\sqrt 2}\right)$$.