How to prove that $( a_n )$ is an increasing sequence by induction?
$n\ge 1$

  • 1
    $\begingroup$ If $n\in\mathbb{R}$, then you don't have a sequence. In fact your definition doesn't even extend to $\mathbb{R}$. $\endgroup$
    – J126
    Oct 24, 2013 at 18:36

1 Answer 1


*(for $n=1$) ${{a}_{2}}=\sqrt{2}>1={{a}_{1}}$ so $a_2>a_1$.

*now assume ${{a}_{n+1}}>{{a}_{n}}.$ we know $\sqrt{{{a}_{n}}+1}={{a}_{n+1}}>{{a}_{n}}$

so using

*${a}_{n+1}>{{a}_{n}}$ and

**$\sqrt{{{a}_{n}}+1}={{a}_{n+1}}$ we get $${{a}_{n+2}}=\sqrt{{{a}_{n+1}}+1}>\sqrt{{{a}_{n}}+1}={{a}_{n+1}}$$ then $${{a}_{n+2}}>{{a}_{n+1}}$$


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