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I found a limit problem which is $\lim_{n\to\infty}{2n \choose n}$ as we know that $\sum_{i=0}^{n}{n \choose i}^2={2n \choose n}$ now we could transform the given problem as $\lim_{n\to\infty} \sum_{i=0}^{n}{n \choose i}^2$ and as $\sum_{i=0}^{n}{n \choose i}^2$ is a polynomial in $n$, therefore, $\lim_{n\to\infty}\sum_{i=0}^{n}{n \choose i}^2=\infty$ Please tell me if my solution is correct or not,and if it is incorrect then please provide me a solution.

Thanks

Anish Ray
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    $\sum_{i=0}^n\binom ni^2$ is as much a polynomial in $n$ as $\binom{2n}n$ is: it is not a fixed polynomial in $n$. So your argument is flawed. However, you can simply say that $$\sum_{i=0}^n\binom ni^2=1+n^2+\text{positive stuff}$$ –  Dec 27 '16 at 12:03
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    Related: https://math.stackexchange.com/questions/58560 – Watson Dec 27 '16 at 12:20
  • https://math.stackexchange.com/questions/1588733/calculate-lim-n-to-infty-binom2nn-without-using-lh%C3%B4pitals-rule – Workaholic Dec 27 '16 at 15:27

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$$\binom{2n}{n} \ge \binom{2n}{1} = 2n \to \infty$$

lhf
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$\binom{2n}{n}$ is called the central Binomial coefficient. Wikipedia says that the Wallis product can be written in the form of an asymptotic for the central Binomial coefficient. That is, $$\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$$

This supposedly has a short proof: $$ {2n \choose n} = \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx < \frac{4^n}{\pi} \int_{-\pi/2}^{\pi/2} e^{-nx^2} dx < \frac{4^n}{\pi} \int_{-\infty}^{\infty} e^{-nx^2} dx = \frac{4^n}{\sqrt{\pi n}}. $$ In the first step, the formula for $\int_{-\pi/2}^{\pi/2} \cos^{2n} x \phantom. dx$ can be proved by using the Beta function.

The third step is clear, and the last step is the well-known Gaussian integral. So we need only justify the the second step.

There we need the inequality $\cos x \leq e^{-x^2/2}$, that is, $$ \log \cos x + \frac{x^2}{2} \leq 0, $$ for $\left|x\right| < \pi/2$, with equality only at $x=0$. This is true because $\log \cos x +\frac12 x^2$ is an even function that vanishes at $x=0$ and whose second derivative ($-\tan^2 x$) is negative for all nonzero $x \in (-\pi/2, \pi/2)$. Hope it helps.

  • A "short" (depending on what you call short) proof could also rely on Stirling approximation. The powers of $n$ and $e$ go away, the power of $4$ in the numerator remains, the $\sqrt 2$ factors cancel out and only one square root of $\pi n$ remains in the denominator. – Patrick Da Silva Dec 27 '16 at 12:31
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Alternatively, $$ f(n)= {2n\choose n} \implies \frac{f(n+1)}{f(n)}\ =\ \frac{2\:(n+1)\:(2n+1)}{\ (n+1)\ (n+1)}\ge2, \quad n \ge0, $$ giving $$ {2n\choose n} \ge 2^n. $$

Olivier Oloa
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Perhaps a more elegant approach: $n^2=\displaystyle{n \choose 1}^2\le\sum_{i=0}^{n}{n \choose i}^2$ and $\displaystyle\lim_{n\to +\infty}n^2=+\infty$.

Fernando Revilla
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$S=\binom{2n}{n}$

$\implies S=\frac{2^n(2n-1)(2n-3)...1}{n!}$

$\implies 2^n<S<\frac{2^n(2n)(2n-2)....2}{n!}$

$\implies 2^n<S<2^{2n}$

Now using sandwich we get the limit as $\infty$

COOLGUY
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$(2n)!/(n! . n!)$ numerator contains one term of denominator and remaining factor of numerator is greater than other term in denominator. And it keeps increasing. So the whole thing tends to Infinity.

jnyan
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