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I would like to know the value of $\displaystyle\lim_{n\rightarrow \infty} \frac{x_n}{n}$, if $\displaystyle\lim_{n\rightarrow \infty}(x_{n+1}-x_n)= c$.

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    Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 2017-02-07
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    depends what $c$ is. – 2017-02-07

3 Answers 3

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Use the Stolz-Cesaro Theorem:

$$\frac{x_{n+1}-x_n}{(n+1)-n}=x_{n+1}-x_n\xrightarrow[n\to\infty]{}c\implies\frac{x_n}n\xrightarrow[n\to\infty]{}c$$

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$$\displaystyle \exists n_0 ,n>n_0 :x_{n+1}-x_{n}\sim c$$ $$x_{n+1}-x_{n}\sim c\\x_{n+2}-x_{n+1}\sim c\\x_{n+3}-x_{n+2}\sim c\\\vdots\\x_{n+k}-x_{n+k-1}\sim c\\ \to x_{n+k}-x_{n}\sim kc \\\to x_{n+k}-x_{n+k-1}\sim c \to x_{n_0+k}\sim x_{n_0}+k c $$so ,If $|x_{n_0}|

$$\displaystyle\lim_{n\rightarrow \infty} \frac{x_n}{n}=\\\displaystyle\lim_{k\rightarrow \infty} \frac{x_{n_0+k}}{k}=\\ \displaystyle\lim_{k\rightarrow \infty} \frac{ x_{n_0}+k c}{k} \to c\\$$

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    Where does the last equation come from? – 2017-02-07
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    @NiklasHebestreit :What do you mean from "last question "? – 2017-02-07
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    Is $x_{n_0} + k = x_{n_0} + kc$? in the last equation you wrote? – 2017-02-07
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Let $c_1N$. It follows that for suitable $a_1,a_2$, we have $a_1+c_1nN$. Hence $\frac{a_1}n+c_1<\frac {x_n}n<\frac{a_2}n+c_2$ for all $n>N$. We conclude that $c_1\le \liminf \frac{x_n}{n}$ and $\limsup \frac{x_n}{n}\le c_2$. As the only restriction was that $c_1