Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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. – Marek Feb 13 '13 at 16:49
@Marek Okay I proved that $\sqrt{1+a_n}-1 \leq a_n$ if $a_n <1$. What do you mean with $O(a_n^2)$ ? – Kasper Feb 13 '13 at 17:19
up vote 2 down vote accepted

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.

share|cite|improve this answer
Aaaah, this prove I do understand, thank you ! – 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.

share|cite|improve this answer
I haven't learned this (limit) comparison test yet. Can you prove this using another test ? – 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 $x\geqslant0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.