Series Made by Limits

Note that

$\begin{aligned} L&=\lim_{n\to\infty}\frac{1^k+2^k+\cdots+n^k}{n^{k+1}} \\&=\lim_{n\to\infty}\frac 1n\sum_{i=1}^n\left(\frac in\right)^k&\small\color{#3D99F6}\text{Using Riemann sums} \\&=\int_0^1x^k\ dx \\&=\frac {x^{k+1}}{k+1}\Bigg|_0^1 \\&=\frac 1{k+1}, \end{aligned}$

we see that the series is actually harmonic series $\sum_{k=1}^\infty\frac 1{k+1},$ which is divergent, making the answer $\boxed{\text{No}}$ .