Prove, using the method of mathematical induction that the following holds true

Question

For natural numbers $n\ge1$ show the following inequality using induction.

$$n^{1/n}\le 1+\sqrt{\frac{2}{n}}$$

Answer 1

To solve this problem, merely binomial theorem will be enough.

The base case $n = 1$ is obviously true. Then we assume that $$n^{\frac1n}\le 1+\sqrt{\frac{2}{n}} \iff n \le \left(1+\sqrt{\frac2n}\right)^n,$$ and we try to prove $$(n+1)^{\frac1{n+1}}\le 1+\sqrt{\frac{2}{n+1}} \iff n+1 \le \left(1+\sqrt{\frac2{n+1}}\right)^{n+1}.$$

$\mathbf{Claim} \quad \left(1+\sqrt{\dfrac2n}\right)^n + 1 \le \left(1+\sqrt{\dfrac2{n+1}}\right)^{n+1}.$

With this claim, it's clear that $n+1 \le \left(1+\sqrt{\dfrac2n}\right)^n+1 \le \left(1+\sqrt{\dfrac2{n+1}}\right)^{n+1}$, so once the claim is proved, we're done.

Proof of the claim

Let $a_n = \left(1+\sqrt{\dfrac2n}\right)^n$. Then we want $a_n + 1\le a_{n+1} \,\forall n \in \Bbb{N}$. This is a bit tricky if we use binomial theorem and condense the higher power terms with a summation sign.

\begin{split} a_n + 1&= \left(1+\sqrt{\frac2n}\right)^n \\ &= 1 + 1 + \sqrt{2n} + \frac{n(n-1)}{2} \frac2n + \sum_{k=3}^n \binom{n}{k} \left(\sqrt{\frac2n}\right)^k \quad \mbox{(Here we assume $n \ge 3$)}\\ &= 2 + \sqrt{2n} + (n-1) + \sum_{k=3}^n \binom{n}{k} \frac{(\sqrt{2n})^k}{n^k} \\ &= 1 + \sqrt{2n} + n + \sum_{k=3}^n \frac{(\sqrt{2n})^k}{k!} \frac{n(n-1)\cdots(n-k+1)}{n^k} \\ &= 1 + \sqrt{2n} + n + \sum_{k=3}^n \frac{(\sqrt{2n})^k}{k!} \frac{n}{n} \frac{n-1}{n} \cdots \frac{n-k+1}{n} \\ &= 1 + \sqrt{2n} + n + \sum_{k=3}^n \frac{(\sqrt{2n})^k}{k!} 1 \left( 1-\frac{1}{n} \right) \cdots \left( 1-\frac{k-1}{n} \right) \\ &< 1 + \sqrt{2(n+1)} + n + \sum_{k=3}^n \frac{(\sqrt{2(n+1)})^k}{k!} 1 \left( 1-\frac{1}{n+1} \right) \cdots \left( 1-\frac{k-1}{n+1} \right) \\ &= 1 + \sqrt{2(n+1)} + \binom{n+1}{2} \frac2{n+1} + \sum_{k=3}^n \frac{(\sqrt{2(n+1)})^k}{k!} \frac{n+1}{n+1} \frac{n}{n+1} \cdots \frac{n-k+2}{n+1} \\ &= 1 + \sqrt{2(n+1)} + \binom{n+1}{2} \frac2{n+1} + \sum_{k=3}^n \binom{n+1}{k} \left(\sqrt{\frac2{n+1}}\right)^k \\ &= \sum_{k=0}^n \binom{n+1}{k} \left(\sqrt{\frac2{n+1}}\right)^k \\ &= \left(1+\sqrt{\frac2n}\right)^n \\ &= a_{n+1} \end{split}

What is left is the base cases for $n = 1$ and $n = 2$.

\begin{split} a_1 + 1 &= 1 + 1 + \sqrt2 < 2 + 2 = 4 = \left( 1 + \sqrt{\frac22} \right)^2 = a_2\\ a_2 + 1 &= 1 + \left( 1 + \sqrt{\frac22} \right)^2 \\ &= 1 + 1 + \sqrt2 \cdot \sqrt2 + \frac22 \\ &= 1 + \sqrt2 \cdot \sqrt2 + \left( \frac22 + 1 \right) \\ &< 1 + \sqrt2 \cdot \sqrt3 + 2 \\ &< 1 + \sqrt2 \cdot \sqrt3 + \binom{3}{2} \frac23 + \left(\sqrt\frac23\right)^3\\ &= \left(1+\sqrt\frac23\right)^3 \\ &= a_3 \end{split}