Let $\lim\limits_{n\to\infty}a_{n+1}-a_{n}=\alpha$. Show that $\lim\limits_{n\to\infty} \frac{a_n}{n}=\alpha$.

Question

Let $\lim\limits_{n\to\infty}a_{n+1}-a_{n}=\alpha$. Show that $\lim\limits_{n\to\infty} \frac{a_n}{n}=\alpha$.

Since $\lim\limits_{n\to\infty}a_{n+1}-a_{n}=\alpha$, we have that for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $$|(a_{n+1}-a_n) - \alpha| < \epsilon \quad \forall n>N.$$ We wish to show that for all $\epsilon>0$ there exists $M\in\mathbb{N}$ such that $$\left|\frac{a_n}{n} - \alpha \right| < \epsilon \quad \forall n>M.$$ So, \begin{align*} \left|\frac{a_n}{n} - \alpha \right| &= \left|\frac{a_n}{n} - \frac{a_{n+1}}{n} + \frac{a_{n+1}}{n} - \alpha \right| \\ &= \left| \frac{1}{n}(a_n - a_{n+1} - \alpha) + \frac{a_{n+1}}{n} + \sum_{j=1}^{n+1} \frac{\alpha}{n} \right| \\ \end{align*} I tried continuing with this pattern, but I did not get anywhere.

Answer 1

I'll turn this into a different exercise.

Suppose that $\{x_n\}$ approaches $x$. Then prove that $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^nx_k=x$$ as well.

If you complete this, your current problem can be addressed. Since $\{a_{n+1}-a_n\}$ approaches $\alpha$, we can conclude $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n(a_{k+1}-a_k)=\alpha$$ The sum however is telescoping. Thus $$\lim_{n\rightarrow\infty}\frac{1}{n}(a_{n+1}-a_1)=\alpha.$$ Algebra of Limits finishes this.

Answer 2

HINT: It is easy to observe that $a_n=C+n\alpha+b_n$ for large $n$, where $(b_n)$ is some sequence tending to 0.

Answer 3

Let $b_n = a_n - a_{n-1}$ then $a_n = a_1 + b_1. + b_2 + ....+ b_n$. Apply Cesaro theorem to. $b_n$.

Answer 4

We have \begin{align} \lim_{n\to \infty}\left|\frac{a_n}{n}-\alpha\right|= & \lim_{n\to \infty}\left|\frac{a_n}{n}-\frac{a_1}{n}-\alpha\right| \\ =& \lim_{n\to \infty}\left|\left(\frac{a_n}{n}-\frac{a_{n-1}}{n}-\frac{\alpha}{n}\right)+\left(\frac{a_{n-1}}{n}-\frac{a_{n-2}}{n}-\frac{\alpha}{n}\right)+\cdots+\left(\frac{a_{2}}{n}-\frac{a_{1}}{n}-\frac{\alpha}{n}\right)\right| \\ =&\lim_{n\to \infty}\left|\frac{(a_{n+1}-a_{n}-\alpha)+(a_{n}-a_{n-1}-\alpha)+\ldots+(a_2-a_1)}{n}\right| \end{align} For all $\epsilon>0$ we there exists $N_0$ such that $n>N_0$ implies $$ |a_{n-1}-a_{n}|<\epsilon. $$ Note that \begin{align} \\ \lim_{n\to \infty}\left|\frac{a_n}{n}-\alpha\right| \leq & \lim_{n\to \infty} \left| \frac{(a_{n+1}-a_{n}-\alpha)+\ldots +(a_{N_0+2}-a_{N_0+1})}{n} \right| \\ +& \lim_{n\to \infty} \left| \frac{(a_{N_0+1}-a_{N_0}-\alpha)+\ldots +(a_{2}-a_{1}-\alpha)}{n} \right| \\ \leq & \lim_{n\to \infty} \frac{|a_{n+1}-a_{n}-\alpha|+\ldots +|a_{N_0+2}-a_{N_0+1}|}{n} \\ +& \lim_{n\to \infty} \left| \frac{(a_{N_0+1}-a_{N_0}-\alpha)+\ldots +(a_{2}-a_{1}-\alpha)}{n} \right| \\ \end{align} Note that quantity $\left| \frac{(a_{N_0+1}-a_{N_0}-\alpha)+\ldots +(a_{2}-a_{1}-\alpha)}{n} \right|$ is limited. Therefore, we have $\lim_{n\to \infty} \left| \frac{(a_{N_0+1}-a_{N_0}-\alpha)+\ldots +(a_{2}-a_{1}-\alpha)}{n} \right|=0$. Therefore, we have \begin{align} \lim_{n\to \infty}\left|\frac{a_n}{n}-\alpha\right| \leq & \lim_{n\to \infty} \frac{|a_{n+1}-a_{n}-\alpha|+\ldots +|a_{N_0+2}-a_{N_0+1}|}{n} \\ \leq & \frac{\overbrace{\epsilon +\ldots +\epsilon}^{n \mbox{times}}}{n} \\ \leq & \epsilon \end{align} As $\epsilon$ is any we can only conclude that $$ \lim_{n\to \infty}\left|\frac{a_n}{n}-\alpha\right|=0 $$

Answer 5

Since $\lim\limits_{n \to \infty}(a_{n+1}-a_n)=\alpha,$ then $\forall \varepsilon>0,\exists N$,when $n>N$,we have $$|a_{n+1}-a_n-\alpha|<\frac{\varepsilon}{2}.$$ Morover, notice that \begin{align*} \left|\frac{a_n}{n}-\alpha\right|&=\left|\frac{a_n-n\alpha}{n}\right|\\ &=\left|\frac{\sum\limits_{k=N+2}^n(a_k-a_{k-1}-\alpha)+\sum\limits_{k=2}^{N+1}(a_{k}-a_{k-1}-\alpha)+(a_1-\alpha)}{n}\right|\\ &\leq \frac{\sum\limits_{k=N+2}^n|a_k-a_{k-1}-\alpha|}{n}+ \frac{\sum\limits_{k=2}^{N+1}|a_{k}-a_{k-1}-\alpha|+|a_1-\alpha|}{n}\\ & \leq \frac{\sum\limits_{k=N+2}^n|a_k-a_{k-1}-\alpha|}{n-N-1}+ \frac{\sum\limits_{k=2}^{N+1}|a_{k}-a_{k-1}-\alpha|+|a_1-\alpha|}{n}\\ &< \frac{\varepsilon}{2}+ \frac{\sum\limits_{k=2}^{N+1}|a_{k}-a_{k-1}-\alpha|+|a_1-\alpha|}{n}. \end{align*} For the part $$ \frac{\sum\limits_{k=2}^{N+1}|a_{k}-a_{k-1}-\alpha|+|a_1-\alpha|}{n},$$ we may always choose a $N'$ such that $$ \frac{\sum\limits_{k=2}^{N+1}|a_{k}-a_{k-1}-\alpha|+|a_1-\alpha|}{n}<\frac{\varepsilon}{2}$$ when $n>N'.$

Thus, we may take $N^*=\max(N,N')$ such that when $n>N^*$,$$\left|\frac{a_n}{n}-\alpha\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$ which implies the limit we want.

Answer 6

Another proof

Denote $$b_n=a_{n+1}-a_n,~~~n=1,2,\cdots.$$ Then $$a_n-a_1=\sum_{k=1}^{n-1}(a_{k+1}-a_{k})=\sum_{k=1}^{n-1} b_k.$$ Hence $$\dfrac{a_n}{n}=\dfrac{a_1+\sum\limits_{k=1}^{n-1} b_k}{n}=\frac{n-1}{n} \left(\frac{a_1}{n-1}+\frac{\sum\limits_{k=1}^{n-1} b_k}{n-1}\right) \to 1\cdot(0+\alpha)=\alpha, ~~~(n \to \infty)$$ where we just cited the fact

$$\lim_{n \to \infty}\frac{x_1+x_2+\cdots+x_n}{n}=\lim_{n \to \infty}x_n.$$

Answer 7

A third proof

By Stolz theorem, $$\lim_{ n \to \infty}\frac{a_n}{n}=\lim_{ n \to \infty}\frac{a_{n+1}-a_n}{(n+1)-n}=\lim_{ n \to \infty}(a_{n+1}-a_n)=\alpha.$$