Another limit related to pi number

Question

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.

Answer 1

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 $t-1/2=u$.

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

$$=2\arctan 1=2\frac{\pi}{4}=\frac{\pi}{2}$$

Answer 2

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

Answer 3

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