Proving that $a_n=\sqrt{n^2+n}-n$ converges, finding its limit and showing that its sequence ${(a_n)^{\infty}_{n=1}}$ is monotinic.

Question

As the title says, below I've proved the statement. Just posted here for verification and correction! And of course for other people to be inspired.

Let $a_n=\sqrt{n^2+n}-n$ with $n\in\mathbb{N}.$ Show that $a_n$ converges as $n\to\infty$ find its limit, and show that the sequence ${(a_n)^{\infty}_{n=1}}$ is monotinic.

Proof:

For convergence, we look for:

$\forall_{\epsilon\gt0}\exists_{n\in\mathbb{N}}\forall_{n\gt N}(|a_n-l|<\epsilon) \iff l-\epsilon<a_n<l+\epsilon.$ If this property holds for one and only one $l$, then $a_n$ convergence with its limit being unique.

$l-\epsilon<\sqrt{n^2+n}-n<l+\epsilon \iff (l+n-\epsilon)^2<n^2+n<(l+n+\epsilon)^2 \iff $

$n(n+l-\epsilon)+l(n+l-\epsilon)-\epsilon(n+l-\epsilon) <n^2+n< $ $n(n+l+\epsilon)+l(n+l+\epsilon)+\epsilon(n+l+\epsilon)$

$\epsilon$ can get randomly small, so it can be ommited for large $n$. Then;

$n(n+l)+l(n+l)<n^2+n< n(n+l)+l(n+l) \iff$ $n^2+2ln+l^2<n^2+n<n^2+2ln+l^2.$

As $n\to\infty$, all terms not containing $n$ can be omitted as they are unimpactfully small. Then;

$n^2+2ln<n^2+n<n^2+2ln \iff 2ln<n<2ln \iff n=2ln \iff l=\frac{1}{2}.$

This property only holds for one $l$, so $a_n$ is convergent, and with this, its limit $l=\frac{1}{2}$ has also been found.

For monotonity we look for:

$\forall_n(\sqrt{(n+1)^2+(n+1)}-(n+1)>\sqrt{n^2+n}-n)$.

For all $n$:

$1>0 \iff 4n^2+4n+1>4n^2+4n \iff (2n+1)^2>4(n^2+n) \iff$ $2n+1>2\sqrt{n^2+n} \iff n^2+2n+1>n^2+2\sqrt{n^2+n} \iff$ $(n+1)^2>n^2+2\sqrt{n^2+n} \iff (n+1)^2+n+1>n^2+n+2\sqrt{n^2+n}+1 \iff$ $\sqrt{(n+1)^2+n+1}>\sqrt{n^2+n}+1 \iff\sqrt{(n+1)^2+(n+1)}-(n+1)>\sqrt{n^2+n}-n$

So the series ${(a_n)^{\infty}_{n=1}}$ is monotonic.

Conclusion:

$a_n$ converges with its unique limit being $l=\frac{1}{2}$, and the series ${(a_n)^{\infty}_{n=1}}$ is monotinic. $\tag*{$\Box$}$

Answer 1

You can "save" your calculation of $l$ like this: you have got $$ n^2+2n(l-\epsilon)+(l-\epsilon)^2<n^2+n<n^2+2n(l+\epsilon)+(l+\epsilon)^2. $$ Removing $n^2$ and dividing by $n$ gives $$ 2(l-\epsilon)+\frac{(l-\epsilon)^2}{n}<1<2(l+\epsilon)+\frac{(l+\epsilon)^2}{n}. $$ Since it has to be true for all large $n$ ($n\to\infty$) it must hold $$ 2(l-\epsilon)\le 1\le 2(l+\epsilon). $$ Since it holds for any $\epsilon>0$ we have $$ 2l\le 1\le 2l. $$

Answer 2

A simpler approach: $a_n>0$,

$$a_n^{-1}=\frac{1}{\sqrt{n^2+n}-n}=\frac{\sqrt{n^2+n}+n}{n}=\sqrt{1+n^{-1}}+1$$

is decreasing and $a_n^{-1}\to 2$. So $(a_n)$ is increasing and $a_n\to 1/2$.

Answer 3

$$a_n={n\over \sqrt{n^2+n}+n}= {1\over \sqrt{1+{1\over n}}+1}$$

so $$ 0<a_n\leq {1\over 2}$$ and since $\sqrt{1+{1\over n}}+1$ is decreasing so is $a_n$ increasing. So $a_n$ is convergent with limit ${1\over 2}$.