Simplify $(n+1)^{1/\sqrt{n}}$ and $(n^2+1)^{1/n}$ Which simplifications can be made to simplify $(n+1)^{1/\sqrt{n}}$ and $(n^2+1)^{1/n}$ to show formally that both expressions converge towards 1 as $n\to\infty$?
 A: If you want to get rid of the annoying $+1$, you can consider that you function is sandwiched between
$$n^{1/\sqrt{n}} < (n+1)^{1/\sqrt{n}} <  (2n)^{1/\sqrt{n}} $$
for $n>1$ and these two bounds should be easier to manage (indeed both have limit $1$).
The other one is similar.
A: I think we can use approximations ie $n+1\approx n$ at large value so the series converges to $1$ same way 2nd series can be approximated.
A: You could tackle it by taking the log and then the limit:
$$\lim_{n\to\infty}\frac{\log(n+1)}{\sqrt{n}}=\lim_{n\to\infty}\frac{2\sqrt{n}}{(n+1)}\tag{L'Hospital's rule}$$
$$=0$$
$$=\log(1)$$
So the limit for the first one is $1$.
$$\lim_{n\to\infty}\frac{\log(n^2+1)}{n}=\lim_{n\to\infty}\frac{2n}{n^2+1}$$
$$=0$$
$$=\log(1)$$
So both limits are $1$.
A: Consider the function, defined for $t>0$,
$$
f(t)=\log\left(\left(\frac{1}{t^2}+1\right)^t\right)
=t\log(1+t^2)-2t\log t
$$
Since $\lim_{t\to0^+}t\log t=0$, we clearly have $\lim_{t\to0^+}f(t)=0$.
Therefore
$$
\lim_{t\to0^+}\left(\frac{1}{t^2}+1\right)^t=
\lim_{t\to0^+}e^{f(t)}=1
$$
Now substitute $t=1/\sqrt{n}$.
The second is the same, with $t=1/n$.
