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I need help to proof next equation: $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n(k(k+1)...(k+m-1))}{\sum_{k=1}^nk^m} = 1$

I know, that $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^nk^m}{{n^{m+1}}} = {\frac{1}{m+1}}$

so $\frac{\sum_{k=1}^n(k(k+1)...(k+m-1))}{n^{m+1}}$ should equals to (m+1)


Stolz–Cesàro theorem $\lim\limits_{n\to\infty}\frac{x_n}{y_n} = \lim\limits_{n\to\infty}\frac{x_n-x_{n-1}}{y_n-y_{n-1}}$

$\lim\limits_{n\to\infty}\frac{n(n+1)...(n+m-1)}{n^m} = \lim\limits_{n\to\infty}\frac{n^m (...)}{n^m} = \lim\limits_{n\to\infty}{(1+1/n)(1+2/n)...(1+(m-1)/n) = 1}$

Awesome! thanks!

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  • $\begingroup$ Hint: divide both sides by $n^m$ and use a Riemann integral. $\endgroup$ – cactus314 Feb 9 '14 at 12:11
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    $\begingroup$ Or use the Cesaro-Stolz theorem, which is similar to l'Hopitals rule in this discrete situation. $\endgroup$ – LutzL Feb 9 '14 at 12:24
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Note that

$$\sum_{k=1}^n k(k+1)...(k+m-1)=m!\sum_{k=1}^n\binom{k+m-1}{m}$$

is a well known binomial sum.

Or use $m+1=(m+k)-(k-1)$ to arrive at a telescoping sum.


And as noted in the comment, the Cesaro-Stolz theorem gives the limit without computing any sums.

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Let $\mathcal{N}_n$ and $\mathcal{D}_n$ be the numerator and denominator in the limit.

For the numerator, we have $$\begin{align} \mathcal{N}_n = \sum_{k=1}^n \prod_{\ell=0}^{m-1}(k+\ell) =&\frac{1}{m+1}\sum_{k=1}^n((k+m)-(k-1))\prod_{\ell=0}^{m-1}(k+\ell)\\ =& \frac{1}{m+1}\sum_{k=1}^n \left(\prod_{\ell=0}^{m}(k+\ell) - \prod_{\ell=0}^{m}(k-1+\ell)\right)\\=& \frac{1}{m+1}\prod_{\ell=0}^m(n+\ell) \end{align} $$ This implies

$$\lim_{n\to\infty} \frac{1}{n^{m+1}} \mathcal{N}_n = \frac{1}{m+1}\lim_{n\to\infty}\prod_{\ell=0}^m\left(1 + \frac{\ell}{n}\right) = \frac{1}{m+1}$$

For the denominator, $$\frac{1}{n^{m+1}}\mathcal{D}_n = \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^m$$ This has the form of a Riemann sum. As a result,

$$\lim_{n\to\infty} \frac{1}{n^{m+1}}\mathcal{D}_n = \int_0^1 x^m dx = \frac{1}{m+1}$$

Combine these two results, we get

$$\lim_{n\to\infty}\frac{\mathcal{N}_n}{\mathcal{D}_n} = \lim_{n\to\infty}\frac{\frac{1}{n^{m+1}}\mathcal{N}_n}{\frac{1}{n^{m+1}}\mathcal{D}_n} = \frac{\frac{1}{m+1}}{\frac{1}{m+1}} = 1$$

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