Can anyone help me with next step? I'm already at this step, how should I change this to solve the problem when $a_n$ is grater than $1$?

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  • $\begingroup$ Welcome to MSE. You can format maths using MathJax. $\endgroup$ – Ennar Nov 4 '18 at 18:24
  • $\begingroup$ $\frac{n-1}{n}\cdot \frac{n+1}{n}=\frac{n^2-1}{n^2}$. You're missing the $^2$ in the denominator in your last expression. $\endgroup$ – Arthur Nov 4 '18 at 18:31
  • $\begingroup$ You will need the inequality by Bernoulli. $\endgroup$ – Dr. Sonnhard Graubner Nov 4 '18 at 18:35
  • $\begingroup$ @Dr.SonnhardGraubner in which point of my equation? $\endgroup$ – Michał Waligóra Nov 4 '18 at 19:10

$$ \frac{a_{n+1}}{a_n}=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}=\frac{n^n(n+2)^{n+1}}{(n+1)^{2n+1}}=\left(\frac{n(n+2)}{(n+1)^2}\right)^n\frac{n+2}{n+1} \\=\left(1-\frac{1}{(n+1)^2}\right)^n\frac{n+2}{n+1}\ge\left(1-\frac{n}{(n+1)^2}\right)\frac{n+2}{n+1}=\frac{(n+2)(n^2+n+1)}{(n+1)^3}=\frac{n^3+3n^2+3n+2}{n^3+3n^2+3n+1}>1. $$ Hence, $\{a_n\}$ is strictly increasing.

Note. We have used the inequality $$ (1+a)^m\ge 1+ma, \quad a>-1, \,\,m\in\mathbb N. $$


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