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Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that the sequence $(a_{n+1}-a_n)_{n \ge 1}$ is convergent to a non-zero real number. Evaluate the limit $$ \lim_{n \to \infty} \left( \frac{a_{n+1}}{a_n} \right)^n.$$

Suppose $\displaystyle\lim_{n \to \infty}(a_{n+1}-a_n)=L>0$. Then given any $\epsilon>0$, there exists some $N\in\mathbb{N}$ such that for all $n\geq\mathbb{N}$, $$L-\epsilon\leq a_{n+1}-a_n\leq L+\epsilon$$. From this how to approach?

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    $\begingroup$ Always a good idea to start with examples. Can you produce any sequence $\{a_n\}$ with the desired property? What is the limit in the examples you can find? Once you have a few examples, you should have a good idea as to the answer, and that is always a big help. $\endgroup$ – lulu May 31 '19 at 13:04
  • $\begingroup$ Nice exercise, anyway. $\endgroup$ – Giuseppe Negro May 31 '19 at 13:11
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Hint: With the same notation as your approach, show that there exists $R\in\mathbb{R}$ such that for all $n\geq N$, you have $$ \lvert a_n-nL-R\rvert\leq n\varepsilon. $$ Hence attack the question along the line $$ \frac{a_{n+1}}{a_n}\approx 1+\frac{1}{n}. $$

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Fix $\epsilon>0$ and pick $N$ accordingly. We may asssume wlog that $\epsilon<\frac L8$ so that $$ 1+\frac1L\epsilon<1+\frac2L\epsilon-\frac8{L^2}\epsilon^2=\left(1+\frac4L\epsilon\right)\left(1-\frac2L\epsilon\right)$$ and hence $$\tag10<\frac{1+\frac1L\epsilon}{1-\frac2L\epsilon}<1+\frac4L\epsilon $$ For $n>N$, we conclude $$(n-N)L-(n-N)\epsilon<a_n-a_N<(n-N)L+(n-N)\epsilon,$$ hence $$ |a_n-nL|<(|a_N-NL|+N\epsilon)+n\epsilon$$ or $$ \left|\frac {a_n}L-n\right|<c+\frac nL\epsilon.$$ For $n\gg 0$ this means $$ \left|\frac {a_n}L-n\right|<\frac {2n}L\epsilon.$$

Thus $$\frac{a_{n+1}}{a_n}=1+\frac{a_{n+1}-a_n}{a_n}<1+\frac{1+ \frac1L\epsilon}{(1-\frac 2L\epsilon)n}<1+\frac{1+\frac4L\epsilon}n $$ and $$ \limsup\left(\frac{a_{n+1}}{a_n}\right)^n\le \lim\left(1+\frac{1+\frac4L\epsilon}{n}\right)^n=e^{1+\frac4L\epsilon}.$$ As this holds for all $\epsilon>0$, we have $$ \limsup\left(\frac{a_{n+1}}{a_n}\right)^n\le e.$$ By the same method, we can show that $$ \liminf\left(\frac{a_{n+1}}{a_n}\right)^n\ge e$$ and conclude $$ \lim\left(\frac{a_{n+1}}{a_n}\right)^n= e.$$

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