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.
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 $t-1/2=u$.
$\frac{1}{2}\int\limits_{ - 1/2}^{1/2} {\frac{{du}}{{{u^2} + {{\left( {1/2} \right)}^2}}}} = \arctan 2\frac{1}{2} - \arctan 2\left( { - \frac{1}{2}} \right)$
$=2\arctan 1=2\frac{\pi}{4}=\frac{\pi}{2}$
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
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 $\sqrt\frac{e \pi}2 \operatorname{erf}\left(\frac{1}{\sqrt 2}\right)$.