# 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.

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?

• 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. – lulu May 31 '19 at 13:04
• Nice exercise, anyway. – Giuseppe Negro May 31 '19 at 13:11

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 $$\frac{a_{n+1}}{a_n}\approx 1+\frac{1}{n}.$$

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 $$(n-N)L-(n-N)\epsilon hence $$|a_n-nL|<(|a_N-NL|+N\epsilon)+n\epsilon$$ or $$\left|\frac {a_n}L-n\right| 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 $$\lim\left(\frac{a_{n+1}}{a_n}\right)^n= e.$$