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I have been studying binomial theorem and have been practicing the methods to find the results of summations involving binomial coefficients and in one of the books I own, I came across this limit of a summation involving binomial coefficients:

$$\lim_{n\to\infty}\sum_{r=0}^n(-1)^r\binom{n}{2r}\left(\frac{x}{n}\right)^{2r}$$

I have tried almost all the methods I have learnt which are to be used in order to find such sums, but this one has completely stumped me. Any kind of hints/solutions/explanations would be highly appreciated.

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    $\begingroup$ $x$ has no role in this. You can just replace it with $1$ $\endgroup$ – polfosol Feb 4 '17 at 8:38
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    $\begingroup$ Please don't vandalize your posts. $\endgroup$ – Glorfindel Feb 4 '17 at 11:54
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HINT: the sum $\sum_{k=0}^n(-1)^r\binom{n}{2k}$ reminds me the series representation of the cosine function and, at the same time, a binomial expansion. Then it seems that this is a binomial expansion of powers of $i$, observe that

$$\sum_{r=0}^n(-1)^r\binom{n}{2r}=\sum_{r=0}^ni^{2r}\binom{n}{2r}$$

and

$$(1+i)^n=\sum_{r=0}^n\binom{n}{r}i^r$$

Hence adding $(x/n)^{2r}$ we can see that

$$(1+ix/n)^n=\sum_{r=0}^n\binom{n}{r}(ix/n)^r\tag{1}$$

and

$$\sum_{r=0}^n(-1)^r(x/n)^{2r}\binom{n}{2r}=\sum_{r=0}^n(ix/n)^{2r}\binom{n}{2r}\tag{2}$$

The key is relate (1) and (2).

Take a look at the binomial expansion of $(1+a)^n+(1-a)^n$

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  • $\begingroup$ @SubhanjanSaha I updated the hint. $\endgroup$ – Masacroso Feb 4 '17 at 10:27
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By exploiting the discrete Fourier transform:

$$ \sum_{r=0}^{n}(-1)^r\binom{n}{2r}\left(\frac{x}{n}\right)^{2r}=\frac{1}{2}\sum_{r=0}^{n}\binom{n}{r}\left[\left(\frac{ix^2}{n^2}\right)^r+\left(\frac{-ix^2}{n^2}\right)^r\right]\tag{1} $$ hence we are looking for: $$ \frac{1}{2}\lim_{n\to +\infty}\left(1+\frac{ix^2}{n^2}\right)^n+\left(1-\frac{ix^2}{n^2}\right)^n = \color{red}{1}\tag{2}$$ since for any $C\in\mathbb{C}$ we have $\lim_{n\to +\infty}\left(1-\frac{C}{n^2}\right)^n=1$.

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