Question regarding the limit of a sequence

You're saying:

1. We write $n = (1+h)^{n+1}$.
2. This necessitates $h<\sqrt{\frac{2}{n+1}}$.
3. Giving you the benefit of the doubt, if we choose $n$ such that $\sqrt{\frac{2}{n+1}} < \epsilon$ (we never choose $\epsilon$ – we must take general $\epsilon>0$), then this shows that $h<\epsilon$.
4. Follow similar steps for $n=(a+h)^{n+1}$.

OK... so we've shown that we can't write $n = a^n$ for $a>1$ as $n$ gets large. I think you'll probably agree this is not what you set out to prove! It certainly doesn't show convergence of the sequence in the question.

What you should be trying to show instead is that $(1+h)^{n+1} \geq n$ for sutiably large $n$ (and for any $h$).

Addition

Does your confusion come from the following? To show the convergence we want, for any $h$, to find $N$ such that for $n>N$ $$ n^\frac{1}{n+1} - 1 < h $$ We then add 1 to both sides to find the condition above. Note that adding $a+1$ to both sides gives $$ n^\frac{1}{n+1} + a < h + 1 + a $$ which is not useful.

You don't choose $\epsilon$. That's given: for every $\epsilon$ you need to find $N$ such that, whenever $n>N$ you have $|h|<\epsilon$.

Your work is good. Since $n>1$, also $n^{1/(1+n)}>1$, so you know that $$ h=n^{1/(1+n)}-1>0 $$ which simplifies things. Since $n=(1+h)^{n+1}$ you can indeed state that $$ n>\frac{n(n+1)h^2}{2} $$ and therefore $$ h<\sqrt{\frac{2}{n+1}} $$ Now take $N$ such that $$ \sqrt{\frac{2}{N+1}}<\epsilon $$ For $n>N$ you have $n+1>N+1$ and so $$ \frac{2}{n+1}<\frac{2}{N+1} $$ and you're done: when $n>N$ you have $$ h<\sqrt{\frac{2}{n+1}}<\sqrt{\frac{2}{N+1}}<\epsilon $$ But can we find $N$? Yes: the condition becomes $$ N+1>\frac{2}{\epsilon^2} $$ and such $N$ certainly exists.

$n\ge 1\implies n^{1/(n+1)}\ge 1$ so it is true that for every $n\ge 1$ there exists $h_n\ge 0$ such that $n^{1/(n+1)}=1+h_n,$ and you deduce correctly that $n^{1/(n+1)}\to 1$.

If you assume that $n\ge 1\implies n^{1/(n+1)}\ge 2$ then you can say that for every $n\ge 1$ there exists $h^*_n\ge 0$ such that $n^{1/(n+1)}=2+h^*_n$ and conclude that $n^{1/(n+1)}\to 2.$

But $n\ge 1\implies n^{1/(n+1)}\ge 2$ is $false$.

And if $n^{1/(n+1)}=2+h^*_n$ with $h^*_n<0$ then the inequalities you wish to obtain by the Binomial Theorem are not valid because the binomial expansion will contain negative and positive terms