Question

Evaluate the limit.


`\displaystyle\rm \lim _(p \rightarrow \infty)\left[\lim _(n \rightarrow \infty)\left(\prod_(k=1)^(n)\left(1+((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 + ((k+1)^p)/(n^(p+1))\right) \\\\ = \exp\left(\lim_(n\to\infty) \ln \prod_(k=1)^n \left(1 + ((k+1)^p)/(n^(p+1))\right)\right) \\\\ = \exp\left(\lim_(n\to\infty) \sum_(k=1)^n \ln \left(1 + ((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(((k+1)^p)/(n^(p+1)) - ((k+1)^(2p))/(n^(2(p+1)))\right) \le \sum_(k=1)^n \ln \left(1 + ((k+1)^p)/(n^(p+1))\right) \le \sum_(k=1)^n ((k+1)^p)/(n^(p+1))`

In the limit, we have

`\displaystyle \lim_(n\to\infty) \sum_(k=1)^n ((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 ((k+1)^(2p))/(n^(2(p+1))) \\\\ = \lim_(n\to\infty) \frac1n * \frac1n \sum_(k=1)^n \left(\frac{k+1}n\right)^(2p) \\\\ = 0 * \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) (H_p)/(p+1)\right) \\\\ = \exp\left(\lim_(p\to\infty) (H_p-\ln(p) + \ln(p))/(p+1)\right) = e^0 = \boxed{1}`

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