Evaluate $\lim_{n\to \infty}\lim_{m\to \infty} \frac{1+\sqrt[n]{1^n+2^n}+\sqrt[n]{2^n+3^n}+\sqrt[n]{3^n+4^n}+.......\sqrt[n]{(m-1)^n+m^n}}{m^2} $

Question

Evaluate the limit

$$\lim_{n\to \infty}\lim_{m\to \infty} \frac{1+\sqrt[n]{1^n+2^n}+\sqrt[n]{2^n+3^n}+\sqrt[n]{3^n+4^n}+.......\sqrt[n]{(m-1)^n+m^n}}{m^2} $$

the answer in the book is $\frac{1}{2}$ but i think the procedure of evaluating mis wrong so plzz solve this!

Answer 1

We can try a double limit version of the squeeze theorem:

$$\sqrt[n]2\frac{(m-1)m}{2m^2}=\sum_{k=0}^{m-1}\frac{\sqrt[n]{2k^n}}{m^2}\le\sum_{k=0}^{m-1}\frac{\sqrt[n]{k^n+(k+1)^n}}{m^2}\le\sum_{k=0}^{m-1}\frac{\sqrt[n]{2(k+1)^n}}{m^2}=\sqrt[n]2\frac{m(m+1)}{2m^2}$$

But

$$\sqrt[n]2\frac{m(m+1)}{2m^2}\xrightarrow[n,m\to\infty]{}1\cdot\frac12=\color{red}{\frac12}\;\;,\;\;\sqrt[n]2\frac{(m-1)m}{2m^2}\xrightarrow[n,m\to\infty]{}1\cdot\frac12=\color{red}{\frac12}$$

Answer 2

There is another solution. By the Stolz–Cesàro theorem, \begin{eqnarray} &&\lim_{n\to \infty}\lim_{m\to \infty} \frac{1+\sqrt[n]{1^n+2^n}+\sqrt[n]{2^n+3^n}+\sqrt[n]{3^n+4^n}+.......\sqrt[n]{(m-1)^n+m^n}}{m^2}\\ &=&\lim_{n\to \infty}\lim_{m\to \infty}\frac{\sqrt[n]{(m-1)^n+m^n}}{m^2-(m-1)^2}\\ &=&\lim_{n\to \infty}\lim_{m\to \infty}\frac{\sqrt[n]{(m-1)^n+m^n}}{2m-1}\\ &=&\frac12. \end{eqnarray}