Prove $(a_n)_{n\in\mathbb{N}}$ with $a_n=\sqrt{n+1}-\sqrt{n}$ to be convergent As the title suggests, I need to prove that the said sequence is convergent. I already assume that it should converge towards 0(it is said to be a real sequence, obviously) but I don't quite know how to perform the epsilon argument. Note: I need to prove it via the core definition of convergence, limit theorems are not allowed.
 A: $\sqrt{n+1}-\sqrt{n}=\sqrt{n}\left(\sqrt{1+\frac 1n}-\sqrt{1}\right)=\underbrace{\dfrac 1{\sqrt{n}}}_{\to 0}\times\underbrace{\dfrac{\sqrt{1+\frac 1n}-\sqrt{1}}{\frac 1n}}_{\to(\sqrt{x})'_{x=1}=\frac 12}\to 0$
$\require{cancel}\sqrt{n+1}-\sqrt{n}=\sqrt{n}\left((1+\frac 1n)^{\frac 12}-1\right)=\sqrt{n}(\cancel{1}+\frac 1{2n}-\cancel{1}+o(\frac 1n))=\frac 1{2\sqrt{n}}+o(\frac 1{\sqrt{n}})\to 0$
$0\le\sqrt{n+1}-\sqrt{n}=\dfrac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\le\dfrac 1{2\sqrt{n}}\to 0$
$\displaystyle 0\le\sqrt{n+1}-\sqrt{n}=\int_n^{n+1}\dfrac{\mathop{dt}}{2\sqrt{t}}\le\int_n^{n+1}\dfrac{\mathop{dt}}{2\sqrt{n}}\le\dfrac{(n+1-n)}{2\sqrt{n}}\le\dfrac 1{2\sqrt{n}}\to 0$
A: Every bounded monotonic sequence is convergent.
1) $a_n =\sqrt{n+1} -√n$, $n\in \mathbb{N}$,
is monotonically decreasing.
Consider: 
$f(x) =(x+1)^{1/2} - x^{1/2}$,  $x>0.$
$f'(x) = (1/2)(x+1)^{-1/2} -(1/2)x^{-1/2} <0.$
Hence $(a_n)_{n\in \mathbb{N}}$ is decreasing.
Lower bound :
Consider:
$n+1 > n ;$  then $\sqrt{n+1} >√n, $
since $g(x)= x^{1/2}$ is strictly increasing.
$\rightarrow$:
$a_n \gt 0.$
$(a_n)_{n\in \mathbb{N}}$ converges
