Convergence Proof Problem $\epsilon, n_0$ proof

Question

I need help proving that for the following sequence it will converge to the given limit $p $using an $\epsilon$ $n_0$ argument (i.e given $\epsilon>0$, determine $n_0$ such that $|p_n - p| < \epsilon$ for all $n \geq n_0$).

The sequence is : $$ p_n = n(\sqrt{1+(1/n)}-1); \quad p = 1/2 $$ I've tried solving the beginning steps in order to figure out what to set $n_0$ greater than as to achieve the inequality. However, I keep coming up with $n(n-1)$ or some variation of that. I'm used to ending up with all the n's in the denominator and it makes it really easy. But I don't know how to handle it when the n's are in the numerator because then you are dealing with the inequality as $n(n-1) < \epsilon$ as what you are trying to prove with the definition.

Answer 1

The inequality you have to manage is $$ \left|n\left(\sqrt{1+\frac{1}{n}}-1\right)-\frac{1}{2}\right|<\varepsilon $$ or $$ \frac{1}{2}-\varepsilon<n\left(\sqrt{1+\frac{1}{n}}-1\right)<\frac{1}{2}+\varepsilon $$ that can be written as $$ \frac{1}{2}-\varepsilon+n<n\sqrt{1+\frac{1}{n}}<\frac{1}{2}+\varepsilon+n $$ Since you can assume $\varepsilon<1/2$, you can square everything.

For instance, the inequality on the right becomes $$ n^2+n<\frac{1}{4}+\varepsilon^2+n^2+\varepsilon+n+2\varepsilon n $$ or $$ 2\varepsilon n>-\frac{1}{4}-\varepsilon^2-\varepsilon $$ that is surely satisfied for every $n$.

Do similarly the inequality on the left.

\begin{gather}\frac{1}{4}+\varepsilon^2+n^2-\varepsilon+n-2\varepsilon n<n^2+n\end{gather}
or \begin{gather}2\varepsilon n>\frac{1}{4}+\varepsilon^2-\varepsilon\end{gather}

Answer 2

Hint: $$\left | \sqrt{n^2+n}-(n+1/2) \right | = \left | \frac{ \left ( \sqrt{n^2+n}-(n+1/2) \right ) \left ( \sqrt{n^2+n} + (n+1/2) \right ) }{\sqrt{n^2+n}+(n+1/2)} \right |$$

I claim that some playing with this argument shows that any $n_0$ with $n_0 \geq 1/(8 \varepsilon)$ will do the job.

You probably did something similar to this to formally find the derivative of $\sqrt{x}$, when you were in calculus.