Let $a_n\in\mathbb{R}_{\geq 0}$ and assume $\sum {a_n}^2$ converges. Show that:

$\sum a_n$ converges $\iff \sum (\sqrt{1+a_n}-1)$ converges

For $\Rightarrow$ I think I need to show that $\sqrt{1+a_n}-1 \leq {a_n}^2 +a_n$. Or something like that ? Any hints ?

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    $\begingroup$ When $a_n$ is small (a necessary condition for the convergence of the sum), $\sqrt{1 + a_n} = 1 + a_n / 2 + O(a_n^2)$. So, up to a sum over $a_n^2$ (which converges), these sums are similar. Your job is to find precise estimates. $\endgroup$ – Marek Feb 13 '13 at 16:49
  • $\begingroup$ @Marek Okay I proved that $\sqrt{1+a_n}-1 \leq a_n$ if $a_n <1$. What do you mean with $O(a_n^2)$ ? $\endgroup$ – Kasper Feb 13 '13 at 17:19

Let $M>0$ be an upper bound of $\{\sqrt{a_n+1}+1\}$. Such an $M$ exists as, $a_n\rightarrow0$ (since $\sum\limits_{n=1}^\infty a_n^2$ converges).

Note that $$ {\sqrt{a_n+1}-1\over a_n} ={1\over\sqrt{a_n+1}+1}. $$ Now, since $a_n\ge0$, for each $n$: $$ {1\over M}\le{1\over\sqrt{a_n+1}+1}\le{1\over 2}. $$ Thus, using the nonnegativity of the $a_n$ again: $$\tag{1} 0\le {a_n\over M}\le{ \sqrt{a_n+1}-1}\le{a_n\over 2}. $$ It follows from $(1)$ and the Comparison Test, that the convergence of $\sum\limits_{n=1}^\infty a_n$ is equivalent to that of $\sum\limits_{n=1}^\infty \bigl(\,\sqrt{a_n+1}-1\,\bigr)$.

Note the hypothesis that $\sum\limits_{n=1}^\infty a_n^2$ is convergent is not needed (and is redundant to boot). The proof above just relies on $(a_n)$ being bounded (and $a_n\ge0$ for each $n$, of course). This is implied if either of the two series $\sum\limits_{n=1}^\infty \bigl(\,\sqrt{a_n+1}-1\,\bigr)$ , $\sum\limits_{n=1}^\infty a_n$ converges.

  • $\begingroup$ Aaaah, this prove I do understand, thank you ! $\endgroup$ – Kasper Feb 13 '13 at 18:45

Using the fact $ \lim a_n =0$ and the comparison test, we have

$$ \lim_{n\to \infty} \frac{ \sqrt{1+a_{n} }-1 }{a_n}=\lim_{n\to \infty} \frac{ 1 } {\sqrt{1+a_{n} }+1 }=\frac{1}{2}>0.$$

Then by the comparison test, either both series converge or both series diverge.

  • $\begingroup$ I haven't learned this (limit) comparison test yet. Can you prove this using another test ? $\endgroup$ – Kasper Feb 13 '13 at 17:17

For every $a_n\geqslant0$, $$\tfrac12a_n-\tfrac18a_n^2\leqslant\sqrt{1+a_n}-1\leqslant\tfrac12a_n.$$ The result follows.

To prove the double inequality above, one studies the variations of the functions $u$ and $v$ defined on $\mathbb R_+$ by $$ u(x)=\sqrt{1+x}-1-\tfrac12x,\qquad v(x)=\sqrt{1+x}-1-\tfrac12x+\tfrac18x^2. $$ For example, $u(0)=0$ and $u'(x)\leqslant0$ for every $x\geqslant0$ hence $u(x)\leqslant0$ for every $x\geqslant0$. Likewise, $v(0)=v'(0)=0$ and $v''(x)\geqslant0$ for every $x\geqslant0$ hence $v(x)\geqslant0$ for every $x\geqslant0$.


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