What is the answer to $n$-th root of $n$ as $n$ approach infinity? Is the answer reach 1?

Question

Is $\sqrt[n]{1+x+x^2+x^3+\dots+x^n} = 1$ when $n = \infty$?

Answer 1

if your goal is to evaluate $\lim_{n\rightarrow\infty}\sqrt[n]{\sum_{k=0}^{n}x^{k}}$ so a standard squeeze theorem trick gives for $x\geq1$

$x=\sqrt[n]{x^{n}}\le\sqrt[n]{\sum_{k=0}^{n}x^{k}}\le\sqrt[n]{(n+1)\cdot x^{n}}=\sqrt[n]{n+1}\cdot x$

and therefore the desired limit equals to $x$. and for $0\le x<1$,

$1=\sqrt[n]{1}\le\sqrt[n]{\sum_{k=0}^{n}x^{k}}\le\sqrt[n]{n+1}{}{\longrightarrow}1$

so the limit is $1$ in that case

Answer 2

This is certainly the case for $|x|<1,$ for then the radicand becomes at $n=+\infty,$ the quantity $$\frac{1}{1-x},$$ so that the root of the quantity expressed as $$(1-x)^{-1/n}=e^{{-1/n}\log(1-x)}$$ goes to $1$ at infinity. This fails for $x=1,$ since then the quantity becomes $(1+n)^{1/n},$ which goes to $e$ at infinity. On the other hand, at $x=-1,$ we have that the quantity becomes, when $n$ is odd, $0,$ and when $n$ is even, $1,$ so that the limit fails to exist in this case. Finally when $x>1,$ the radicand being comparable to $x^{n+1},$ has an $n$th root which goes to something different from $1.$