$\lim_{n \to \infty} \sqrt[n] {\frac{2^{\left(\frac{n^2+n+1}{n}\right)}-1}{n^2+n+1}}$

Question

I need to compute the following limit: $$\lim_{n \to \infty} \sqrt[n] {\frac{2^{\left(\frac{n^2+n+1}{n}\right)}-1}{n^2+n+1}}$$

I was solving a problem about the limit of an integral and I arrived after a lot of algebraic manipulation to this expression. I don't know if it's possible to simplify more, and also I don't know how to compute it. I only know that the answer is $2$ (by W.A). I was thinking of squeezing the expression below the $n$-th root, but I couldn't find good bounds either.

Any hints are appreciated

Answer 1

$$\lim_{n \to \infty} \sqrt[n] {\dfrac{2^{\left(\frac{n^2+n+1}{n}\right)}-1}{n^2+n+1}} = \lim_{n\to\infty}\left(\dfrac{2^{\left(\frac{n^2+n+1}{n}\right)}-1}{2^{\left(\frac{n^2+n+1}{n}\right)}}\right)^\frac{1}{n}\times\lim_{n\to\infty}\left(\dfrac{n^2}{n^2+n+1}\right)^\frac{1}{n}\times\dfrac{\lim_{n\to\infty}2^{\frac{n^2+n+1}{n^2}}}{\lim_{n\to\infty}n^\frac2n}\\ =1\times 1\times \dfrac21=\boxed2$$

Answer 2

Pick $\epsilon>0$, $\exists N \in \mathcal N$ s.t. $\forall n>N$: $(2-\epsilon）^n<\frac{2^{\left(\frac{n^2+n+1}{n}\right)}-1}{n^2+n+1}<(2-\epsilon)^n$ because it's clear that $\frac{2^{\left(\frac{n^2+n+1}{n}\right)}-1}{n^2+n+1}\to2^n$.

Then it should be very easy to see that the limit is 2.

Answer 3

In these cases, using asymptotics can be very powerful. If you are not familiar with it, it consists of defining an equivalence relation $\sim$ such that $f(x) \sim g(x) \iff \lim_{x\to\infty} f(x)/g(x) = 1$. First, see if you can show that $\lim_{n\to\infty} (n^2+n+1)^{1/n} = 1$. We also know that $2^{n+1+1/n}-1 \sim 2^{n+1+1/n}$.Your limit then reduced to $$\lim_{n\to\infty}2^{1+1/n+1/n^2} = 2.$$