I was asked to prove the following : \begin{array}{l} x_n > 0,\,\forall n \in Z^ + \\ \lim \sup (\frac{{x_{n + 1} + x_1 }}{{x_n }})^n \ge e \\ \end{array} Is my approach correct ? Using reductio ad absurdum I intend to prove that for any natural number $k$ there is a natural number $n$ such that $\ n \ge k,\,\frac{{x_{n + 1} + x_1 }}{{x_n }} \ge 1 + \frac{1}{n} \ $ . We hereby assume the contrary, namely for any natural number $k$ there is a natural number $n$ satisfying : $\ \frac{{x_{n + 1} + x_1 }}{{x_n }} < 1 + \frac{1}{n} \ $. Then $\ \frac{{x_n }}{n} - \frac{{x_{n + 1} }}{{n + 1}} > \frac{{x_1 }}{{n + 1}} \ $ . Upon summation for $\ n = k,k + 1,...,k + s \ $ we get :
$$\ \frac{{x_k }}{k} - \frac{{x_{k + s} }}{{k + s}} > x_1 (\frac{1}{{k + 1}} + \frac{1}{{k + 2}} + ... + \frac{1}{{k + s}}) \ $$
Since the RHS is unbounded we conclude that the contrary is true :$\ \begin{array}{l} n \ge k,\,\frac{{x_{n + 1} + x_1 }}{{x_n }} \ge 1 + \frac{1}{n} = > \\ = > \lim \sup (\frac{{x_{n + 1} + x_1 }}{{x_n }})^n \ge \lim \sup (1 + \frac{1}{n})^n \\ \end{array} \ $
Hence $$\ lim\sup (\frac{{x_{n + 1} + x_1 }}{{x_n }})^n \ge e \ $$
Some of my peers suggested that the inequality in the box might be wrong in several cases, depending on our choosing of $x_n$ - e.g. $x_n=n^5$ - but I feel that it should hold no matter how $x_n$ looks like (with respect tot the original condition).
Is my proof correct ? Moreover, if my inequality holds for any $x_n>0$ how can I give a formal proof of this fact ?