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((\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((\ln(\pi))/(k^2+k)\right)}{\frac1{k^2}}`

Applying L'Hopital's rule yields

`\displaystyle \lim_(k\to\infty) \frac{\ln(\pi) (2k+1)/((k^2+k)^2) \exp\left((\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)}`