How to show by the Root Test that $\sum\limits_{i=1}^\infty (2n^{1/n}+1)^n$ converges or diverges

Question

How do I show by the Root Test that $$\sum\limits_{i=1}^\infty (2n^{1/n}+1)^n$$ converges or diverges? This is what I have done so far. Since we take $\sum\limits_{i=1}^\infty \sqrt[n]{|a_n|}$, we let $a_n = (2n^{1/n}+1)^n.$ This yields $\sum\limits_{n=1}^\infty\sqrt[n]{(2n^{1/n}+1)^n}$, which simplifies to $\sum\limits_{n=1}^\infty 2n^{1/n}+1.$ I know that $2n^{1/n}$ is an indeterminate form in the form of ${\infty}^0$, which I can solve accordingly. However, what do I do with the $1$? Can I disregard it since $x \rightarrow \infty$ and the $1$ becomes insignificant? That's where I'm stuck on.

Answer 1

Note that $$ 2n^{1/n}+1\ge 3, $$ and thus $$ (2n^{1/n}+1)^n\ge 3^n\ge 3, $$ which implies that the series $\sum_{n=1}^\infty (2n^{1/n}+1)^n$ diverges to infinity, because if a series $\sum_{n=1}^\infty a_n$ converges, then $a_n\to 0$.

Note. If instead we had $\sum_{n=0}^\infty(2n^{1/n}-1)^n$, we would still have divergence, as $(2n^{1/n}-1)^n\ge n$.

Answer 2

By the root test you get

$$\limsup\sqrt[n]{\left|2n^{1/n}+1\right|^n} = \limsup 2n^{1/n}+1 = 3 \gt 1$$

Thus your series diverges.