We have
\[ \lim_{n\to\infty} \big(1 + \tfrac{1}{n}\big)^n \; = \; e \]
and we can also show that the sequence \(\big(1+\tfrac{1}{n}\big)^n\) is an increasing one. Thus we deduce that \(\big(1+\tfrac{1}{n}\big)^n \le e \le n\) for all \(n \ge 3\). Thus
\[ \begin{array}{rcl}
\ln n & \ge & n\ln\big(1+ \tfrac{1}{n}\big) \\
(n+1) \ln n & \ge & n\ln(n+1) \\
\tfrac{1}{n}\ln n & \ge & \tfrac{1}{n+1}\ln(n+1)
\end{array} \]
for all \(n \ge 3\). This tells us that \(n^{\frac{1}{n}} \, \ge \, (n+1)^{\frac{1}{n+1}}\) for \(n \ge 3\).

Since \(1 < 2^{\frac12} < 3^{\frac13}\), we deduce that \(3^{\frac13}\) is the largest value in the sequence.
–
Mark Hennings
·
3 years, 10 months ago

## Comments

Sort by:

TopNewestWe have \[ \lim_{n\to\infty} \big(1 + \tfrac{1}{n}\big)^n \; = \; e \] and we can also show that the sequence \(\big(1+\tfrac{1}{n}\big)^n\) is an increasing one. Thus we deduce that \(\big(1+\tfrac{1}{n}\big)^n \le e \le n\) for all \(n \ge 3\). Thus \[ \begin{array}{rcl} \ln n & \ge & n\ln\big(1+ \tfrac{1}{n}\big) \\ (n+1) \ln n & \ge & n\ln(n+1) \\ \tfrac{1}{n}\ln n & \ge & \tfrac{1}{n+1}\ln(n+1) \end{array} \] for all \(n \ge 3\). This tells us that \(n^{\frac{1}{n}} \, \ge \, (n+1)^{\frac{1}{n+1}}\) for \(n \ge 3\).

Since \(1 < 2^{\frac12} < 3^{\frac13}\), we deduce that \(3^{\frac13}\) is the largest value in the sequence. – Mark Hennings · 3 years, 10 months ago

Log in to reply