Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\{X_n\}$ be a sequence such that:

$$\lim\limits_{n \to \infty } ({X_{n + 1}} - {X_n}) = c$$

Prove that:

$$\lim\limits_{n \to \infty } {{{X_n}} \over n} = c$$

I've tried many approaches here, but not sure how to connect between the two limits.
Any help will be appreciated.

share|cite|improve this question
Hint: $X_n=X_1+[X_{n}-X_{n-1}]+[X_{n-1}-X_{n-2}]+\cdots$ – Alex R. Nov 7 '13 at 22:45
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. – Lord_Farin Nov 7 '13 at 22:47
Hey @AlexR. can you explain how did you come up with this series? or why this equality is true? – captain dragon Nov 8 '13 at 9:30
up vote 5 down vote accepted

Fix $\epsilon > 0$.

Limsup: Choose $N$ large so that $X_n - X_{n-1} \le c + \epsilon$ for all $n \ge N$.

Then choose $M$ large enough so that $\frac{x_N}{M} \le \epsilon$.

for $n \ge \max(M,N)$ write

$$ x_n = x_N + \left( x_{N+1} - x_{N}\right) + \cdots + \left( x_n - x_{n-1} \right)$$

Then $$\frac{x_n}{n} = \frac{x_{N} + \left( x_{N+1} - x_N\right) + \cdots + \left( x_n - x_{n-1} \right)}{n}$$

Hence $\frac{x_n}{n} \le \epsilon + \frac{n - N}{n}(c + \epsilon)$.

Taking $n\rightarrow\infty$ we conclude that $\limsup \frac{x_n}{n} \le c + 2\epsilon$.

So since $\epsilon$ is arbitrary we conclude that $\limsup \frac{x_n}{n} \le c$.

Liminf: This time choose $N$ large so that $X_n - X_{n-1} \ge c - \epsilon$ for $n \ge N$.

Choose $M$ large so that $\frac{x_N}{M} \ge -\epsilon$. Now run through the same argument as before to conclude that $$\frac{x_n}{n} \ge - \epsilon + \frac{n - N}{n} (c - \epsilon)$$

taking $n \rightarrow \infty$ and conclude that $\liminf \frac{x_n}{n} \ge c - 2\epsilon$. since $\epsilon$ was arbitrary we get $\liminf \frac{x_n}{n} \ge c$.

Combining these we conclude that $\lim \frac{x_n}{n} = c$.

share|cite|improve this answer

Let $\log y_{n} = X_{n}$ and then we get $$\lim_{n \to \infty}\log\left(\frac{y_{n + 1}}{y_{n}}\right) = c$$ so that $\lim_{n \to \infty}y_{n + 1}/y_{n} = e^{c} > 0$ and hence $\lim_{n \to \infty} y_{n}^{1/n} = e^{c}$ and taking logs we get $$\lim_{n \to \infty}\frac{X_{n}}{n} = c$$

share|cite|improve this answer
It can also be proved by setting $a_{n} = X_{n + 1} - X_{n}$ so that $a_{1} + a_{2} + \cdots + a_{n} = X_{n + 1} - X_{1}$. Since $a_{n} \to c$ as $n \to \infty$ we get $\sum_{k = 1}^{n}a_{k}/n \to c$ as $n \to \infty$. This means that $\{X_{n + 1} - X_{1}\}/n \to c$ as $n \to \infty$. – Paramanand Singh Nov 8 '13 at 14:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.