Question

$\rm \lim_{k \to \infty } {k}^{2} ( \sqrt[ k + 1]{\pi} - \sqrt[k]{\pi} )$​

Answer 1

Rewrite the limand as

$k^2 \left(\pi^{\frac1{k+1}} - \pi^{\frac1k}\right) = k^2 \pi^{\frac1{k+1}} \left(1 - \pi^{\frac1{k^2+k}}\right) = k^2 \pi^{\frac1{k+1}} \left(1 - \exp\left(\frac{\ln(\pi)}{k^2+k}\right)\right)$

In the limit, the $\pi^{1/(k+1)}$ term converges to 0, and the rest we can rearrange as

$\displaystyle \lim_{k\to\infty} \frac{1 - \exp\left(\frac{\ln(\pi)}{k^2+k}\right)}{\frac1{k^2}}$

Applying L'Hopital's rule yields

$\displaystyle \lim_{k\to\infty} \frac{\ln(\pi) \frac{2k+1}{(k^2+k)^2} \exp\left(\frac{\ln(\pi)}{k^2+k}\right)}{-\frac2{k^3}}$

The exponential term converges to 0 and the rational expression converges to 1, so the overall limit is $\boxed{-\ln(\pi)}$