Question

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


Where, Hp is the harmonic number.​

Answer 1

Rewrite the inner limit as

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

For any x > 0, we have the useful bounds

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

so that

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

In the limit, we have

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

and

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

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

Finally,

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

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