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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.

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2  
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
1  
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

2 Answers 2

up vote 4 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 $\lim \frac{x_n}{n} = c$.

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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 $$\lim_{n \to \infty}\frac{X_{n}}{n} = c$$

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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

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