Proof that the limit as n approaches infinity of n^1/n = 1 (\(\lim_{n \to \infty} n^{1/n} = 1\))

\(\)Here’s an important limit from real analysis that gives quite a few people, including myself, a lot of trouble:
$$
\lim_{n \to \infty}n^{1/n} = 1
$$

Here is the proof that my Advanced Calculus professor at Western Michigan University, Yuri Ledyaev, gave in class. It uses the binomial expansion.

Proof: Since \(n \in \mathbb{N} \), for all \(n \geq 2\) we can write
$$
\begin{eqnarray}
n^{1/n} &=& 1 + \alpha ~ [where ~ \alpha \geq 0] \nonumber \\
(n^{1/n})^n &=& (1 + \alpha)^n \nonumber \\
n &=& (1 + \alpha)^n \nonumber \\
\end{eqnarray}
$$

We want to estimate \( \alpha \). If \( \alpha\) is, say, \(0\), then we’ll have \(n^{1/n} = 1+0\), meaning the limit we’re after will be \(1\). The binomial theorem says that
$$
\begin{eqnarray}
(a+b)^n &=& a^n + na^{n-1}b^1 + \frac{n(n-1)}{2}a^{n-2}b^2 + … +b^n \nonumber \\
&=& \sum\limits_{k=0}^n \binom{n}{k} a^{n-k}b^k\nonumber \\
\end{eqnarray}
$$
Therefore,
$$
\begin{eqnarray}
(1+\alpha)^n &=& 1^n + \binom{n}{1}1^{n-1}\alpha + \binom{n}{2}1^{n-2}\alpha^2 + … +\alpha^n \\[3pt] \nonumber \\
&=& 1^n + n \cdot 1 \cdot \alpha + \frac{n(n-1)}{2} \cdot 1 \cdot \alpha^2 + … + \alpha^n\\[3pt] \nonumber \\
&=& 1 + n\alpha + \frac{n(n-1)}{2}\alpha^2 + … +\alpha^n \\[3pt] \nonumber \\
&>& 1 + n\alpha + \frac{n(n-1)}{2}\alpha^2 \\[3pt] \nonumber \\
&>& 1 + \frac{n(n-1)}{2}\alpha^2 \nonumber \\
\end{eqnarray}
$$

So we have
$$
\begin{eqnarray}
1+\frac{n(n-1)}{2}\alpha^2 &<& (1+\alpha)^n = n \\[3pt] \nonumber \\ \frac{n(n-1)}{2}\alpha^2 &<& n - 1 < n \\[3pt] \nonumber \\ \alpha^2 &<& \frac{n}{\frac{n(n-1)}{2}} \\[3pt] \nonumber \\ \alpha^2 &<& \frac{2}{n-1} \\[3pt] \nonumber \\ \alpha &<& \sqrt{\frac{2}{n-1}} \nonumber \\ \end{eqnarray} $$ Thus, \(\lim_{n \to \infty}\alpha = 0 \), and \(\lim_{n \to \infty}n^{1/n} = \lim_{n \to \infty}(1+\alpha) = 1+0=1\). \(\blacksquare\)

This entry was posted in Math. Bookmark the permalink.

One Response to Proof that the limit as n approaches infinity of n^1/n = 1 (\(\lim_{n \to \infty} n^{1/n} = 1\))

  1. Alex says:

    This is absolutely brilliant, and it helps me to prove that (n!)^(1/n) < (n+1)/2 by induction.
    Thanks for posting