greatest series

which term of the following series is largest 1, 2^(1/2) , 3^(1/3) ....................... , n^(1/n).

#MathProblem

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 - 7 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$