For  $$\alpha \geq 0$$​ , define

​$$a_n = \frac{1 + 2^\alpha + \cdots + n^\alpha}{n^{\alpha+1}}.$$​​

What is the value of  $$\lim\limits_{n \to \infty} a_n$$​ ?

Correct option is C

$$\mathbf{\frac{1}{\alpha + 1}}$$​

Using the Stolz-Cesàro theorem,

​$$\lim_{n\to\infty}\frac{x_n}{y_n}=\lim_{n\to \infty}\frac{x_{n+1}-x_n}{y_{n+1}-y_n} \textit{ Where } \ y_n \textit{ is an increasing sequence}$$​​

we evaluate the given sequence:

​$$\lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} \frac{1 + 2^\alpha + 3^\alpha + \dots + n^\alpha}{n^{\alpha+1} + 1}$$​.


Applying the theorem,

​$$\lim\limits_{n \to \infty} \frac{(n+1)^\alpha - n^\alpha}{(n+1)^{\alpha+1} - n^{\alpha+1}}$$​​


which simplifies to:
​$$\frac{1}{\alpha + 1}.$$​​

​$$\implies \textbf{Option (C) is correct.}$$​​