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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.

user 1591719
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3 Answers3

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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 $t-1/2=u$.

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

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

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Pedro
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  • Does this qualify as "only...using high school knowledge"? – Gerry Myerson May 29 '12 at 01:09
  • I had read the question a long time ago. I really forgot about that. Thanks for pointing it out. I'll try to find a solution using highschool knowledge, but it'd be good if the OP explicitly gives a repertoire. Some people learn definite integrals at highschool. The geometric series is easily taughtable, and the Beta integral might be a harder thing to get through. – Pedro May 29 '12 at 01:12
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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

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

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murray
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  • i think that the limit is $\frac {\pi}2$. – user 1591719 May 27 '12 at 21:16
  • @Peter Tamaroff: is that result equal to $\frac {\pi}2$? – user 1591719 May 27 '12 at 21:18
  • No clue, I haven't checked it. I merely edited. – Pedro May 27 '12 at 21:20
  • @ Peter Tamaroff: i thought you edited the result because you knew the answer. I know what is its limit and that's why i was asking for. As regards the limit, it is definitely $\frac {\pi}2$. – user 1591719 May 27 '12 at 21:24
  • @Chris I just TeXified it. I know the limit is $\pi /2$ but maybe murray can explain himself. – Pedro May 27 '12 at 21:55
  • murray, the sum in question was $$\sum_{k=0}^\infty \frac{{k!}^{2} {2}^{k}}{(2k+1)!} = \frac{\pi}{2},$$ not $$\sum_{k=0}^\infty \frac{k! {2}^{k}}{(2k+1)!} = \sqrt\frac{e \pi}2 \operatorname{erf}\left(\frac{1}{\sqrt 2}\right).$$ Note the square in the numerators of the summands. – Antonio Vargas May 27 '12 at 22:27
  • Sorry, missed that square on the $k!$. – murray May 28 '12 at 03:34
  • According to Mathematica, the tail-end of the infinite series -- after the terms through $k = n$, sums to a rational function of factorials and the hypergeometric $_2F_1(1, n + 2, n + 5/2; 1/2)$. So my answer to the original question would still be no -- at least assuming there's no simplification that avoids the hypergeometric function or some combination of gamma function expressions. – murray May 28 '12 at 03:38