Question

Evaluate the limit.
$\displaystyle\lim _{p \rightarrow \infty}\left[\lim _{n \rightarrow \infty}\left(\prod_{k=1}^{n}\left(1+\frac{(k+1)^{p}}{n^{p+1}}\right)\right)\right]^{H_{p}}$

Where, Hp is the harmonic number.

Answer 1

Rewrite the inner limit as

$\begin{gathered}\displaystyle \rm\lim_{n\to\infty} \prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) \\\\ \rm= \exp\left(\lim_{n\to\infty} \ln \prod_{k=1}^n \left(1 + \frac{(k+1)^p}{n^{p+1}}\right)\right) \\\\ \rm= \exp\left(\lim_{n\to\infty} \sum_{k=1}^n \ln \left(1 + \frac{(k+1)^p}{n^{p+1}}\right)\right)\end{gathered}$

For any x > 0, we have the useful bounds

$\rm x - x^2 \le \ln(1 + x) \le x$

so that

$\displaystyle \rm \sum_{k=1}^n \left(\frac{(k+1)^p}{n^{p+1}} - \frac{(k+1)^{2p}}{n^{2(p+1)}}\right) \le \sum_{k=1}^n \ln \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) \le \sum_{k=1}^n \frac{(k+1)^p}{n^{p+1}}$

In the limit, we have

$\begin{gathered}\displaystyle \rm\lim_{n\to\infty} \sum_{k=1}^n \frac{(k+1)^p}{n^{p+1}} \\\\ \rm= \lim_{n\to\infty} \frac1n \sum_{k=1}^n \left(\frac{k+1}n\right)^p \\\\ \rm= \int_0^1 x^p \, dx = \frac1{p+1}\end{gathered}$

and

$\begin{gathered}\displaystyle \rm\lim_{n\to\infty} \sum_{k=1}^n \frac{(k+1)^{2p}}{n^{2(p+1)}} \\\\ \rm= \lim_{n\to\infty} \frac1n \times \frac1n \sum_{k=1}^n \left(\frac{k+1}n\right)^{2p} \\\\ = \rm0 \times \int_0^1 x^{2p} \, dx = 0\end{gathered}$

Then by the squeeze theorem, the log sum converges to $\rm \frac{1}{p + 1}$ , so the infinite product converges to $\rm e^{\frac1{p+1}}$ , as suspected.

Finally,

$\begin{gathered}\displaystyle \rm\lim_{p\to\infty} \left(e^{\frac1{p+1}}\right)^{H_p} \\\\ = \rm \exp\left(\lim_{p\to\infty} \frac{H_p}{p+1}\right) \\\\ = \rm\exp\left(\lim_{p\to\infty} \frac{H_p-\ln(p) + \ln(p)}{p+1}\right) = e^0 = {1}\end{gathered}$

since Hₙ - ln(n) converges to the Euler-Mascheroni constant γ.

$\displaystyle\rm\lim _{p \rightarrow \infty}\left[\lim _{n \rightarrow \infty}\left(\prod_{k=1}^{n}\left(1+\frac{(k+1)^{p}}{n^{p+1}}\right)\right)\right]^{H_{p}}=1$

Answer 2

Answer:

$a−x2≤ln(1+x)≤x</p><p></p><p>so that</p><p></p><p>\displaystyle \rm \sum_{k=1}^n \left(\frac{(k+1)^p}{n^{p+1}} - \frac{(k+1)^{2p}}{n^{2(p+1)}}\right) \le \sum_{k=1}^n \ln \left(1 + \frac{(k+1)^p}{n^{p+1}}\right) \le \sum_{k=1}^n \frac{(k+1)^p}{n^{p+1}}k=1∑n(np+1(k+1)p−n2(p+1)(k+1)2p)≤k=1∑nln(1+np+1(k+1)p)≤k=1∑nnp+1(k+1)p</p><p></p><p>In the limit, we have</p><p></p><p>\begin{gathered}\begin{gathered}\displaystyle \rm\lim_{n\to\infty} \sum_{k=1}^n \frac{(k+1)^p}{n^{p+1}} \\\\ \rm= \lim_{n\to\infty} \frac1n \sum_{k=1}^n \left(\frac{k+1}n\right)^p \\\\ \rm= \int_0^1 x^p \, dx = \frac1{p+1}\end{gathered}\end{gathered}n→∞limk=1∑nnp+1(k+1)p=n→∞limn1k=1∑n(nk+1)p=∫01xpdx=p+11</p><p></p><p>$