A calculus problem by Jose Sacramento

The limit can be found using Stolz–Cesàro theorem as follows.

Theorem 1 (the $\infty/\infty$ case) :
Let $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ be two sequences of real numbers such that
(a) $0<b_1 < b_2 < \cdots < b_n < \cdots$ and $\displaystyle \lim_{n\to\infty} b_n = \infty$ .
(b) $\displaystyle \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} = l \in \mathbb R$ .

Then, $\displaystyle \lim_{n\to\infty} \dfrac{a_n}{b_n}$ exists and is equal to $l$ .

In this case, $\displaystyle a_n = \sum_{k=1}^n \frac k{k+1}$ and $b_n = n$ . Therefore,

$\begin{aligned} \lim_{n\to\infty} \dfrac{a_{n+1} - a_n}{b_{n+1} - b_n} & = \lim_{n\to\infty} \frac {\sum_{k=1}^{n+1} \frac k{k+1} - \sum_{k=1}^n \frac k{k+1}}{(n+1)-n} \\ & = \lim_{n\to\infty} \frac {\frac {n+1}{n+2}}{1} \\ & = \lim_{n\to\infty} \frac {n+1}{n+2} \\ & = \lim_{n\to\infty} \frac {1+\frac 1n}{1+\frac 2n} \\ & = \boxed{1} \end{aligned}$

Generalisation of this problem:

Proposition.- If the sequence $(a_n)_{n =1}^{\infty}$ fulfills $\displaystyle \lim_{n\to \infty} a_n = a$ where $a$ is a finite number, then $\large \displaystyle \lim_{n\to \infty} \frac{\sum_{k =1}^{n} a_k}{n} = a$ Sketch of Proof.- Applying Stolz-Cesaro Theorem if $a \neq 0$ . If $a = 0$ it is "trivial" q.e.d $\square \\ \quad$

In this case, $\lim_{k\to \infty} \frac{k}{k+1} = 1 \Rightarrow$ $\large \displaystyle \lim_{n\to \infty} \frac{\sum_{k =1}^{n} \frac{k}{k+1}}{n} = 1$

Other example.- $\large \displaystyle \lim_{n\to \infty} \frac{\sum_{k =1}^{n} \frac{2k^2}{(k+1)^2}}{n} = 2$

The sum will become a large number added by infinity The n will be infinity Those two devided by one other Gives 1