Find the limit and show that it converges to that limit. Consider the sequence $\{(\sqrt{n+1}-\sqrt{n})\sqrt{n}\}$. Find the limit and prove it converges to that limit or prove that it does not converge.
My attempt:
$(\sqrt{n+1}-\sqrt{n})\sqrt{n}=\sqrt{n}\sqrt{n+1}-n=\sqrt{n^2+n}-n$
Multiplying by the conjugate then dividing by it gives:
$\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{\frac{n+1}{n}}+1}$.
Then taking the limit of this gives 1/2.
To prove that the sequence converges to 1/2, I have to show that $|(\sqrt{n+1}-\sqrt{n})\sqrt{n}-\frac{1}{2}|&lt;\epsilon$ for some $\epsilon&gt;0$. I know that I have to find an index $N&lt;\epsilon$ but that is where I am stuck.
Just work with the equivalent expression which you found: $\frac{1}{\sqrt{\frac{n+1}{n}}+1}$
Take an arbitrary $\epsilon$
Find N such that
$|\frac{1}{\sqrt{\frac{n+1}{n}}+1} - \frac{1}{2}|&lt;\epsilon$
when $n \gt N$
Hint:  Can you use the binomial theorem?
\begin{aligned}\sqrt{n}\big(\sqrt{n+1}-\sqrt n\big) &amp;= n\Bigg(\Big(1 + \frac{1}{n}\Big)^{1/2}-1\Bigg) \\
&amp;=\Bigg(\frac{1}{2}-\frac{1}{8n}+\cdots\Bigg)
\end{aligned}
Gives a handle on the error term and therefore a route to find $N$ needed to make the error less than $\varepsilon$