$\sum a_n$ converges $\iff \sum (\sqrt{1+a_n}-1)$ converges 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 ?
 A: 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.
A: 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. 
A: 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 $x\geqslant0$.
