Prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals.

Question

I need to prove that $\sqrt[n]{n} < (1 + \frac{1}{\sqrt{n}})^2$ for all n in the naturals. I started by using Bernoulli's inequality: $(1+\frac{2}{\sqrt{n}}) < (1 + \frac{1}{\sqrt{n}})^2$ I can say that: $(1+\frac{2}{\sqrt{n}}) = (1+\frac{2\sqrt{n}}{n})$ I can also subtract the one and divide by 2 on the left side without changing the inequality (because it makes it even smaller): $(\frac{\sqrt{n}}{n}) < (1 + \frac{1}{\sqrt{n}})^2$ But now I am stuck...

Answer 1

Inequality Requested $$ \begin{align} \left(1+\frac1{\sqrt{n}}\right)^{2n} &\ge\left(1+\sqrt{n}\right)^2\tag1\\ &\ge n\tag2 \end{align} $$ Explanation:
$(1)$: Bernoulli's Inequality
$(2)$: $1+\sqrt{n}\ge\sqrt{n}$

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Slightly Stronger Inequality $$ \begin{align} \left(1+\sqrt{\frac2n}\right)^n &\ge1+n\sqrt{\frac2n}+\frac{n(n-1)}2\frac2n\tag3\\ &\ge n\tag4 \end{align} $$ Explanation:
$(3)$: Binomial Theorem
$(4)$: $n+\sqrt{2n}\ge n$

Answer 2

Beware: overkill incoming. We may use the AM-GM inequality for producing tight bounds for $\sqrt[n]{n}$.
In particular, by noticing that $$ n = 1\cdot \frac{2}{1}\cdot\frac{3}{2}\cdot\ldots\cdot\frac{n}{n-1} $$ we have $$ \sqrt[n]{n}=\text{GM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right)<\text{AM}\left(1,1+1,1+\frac{1}{2},\ldots,1+\frac{1}{n-1}\right) $$ and $$ \sqrt[n]{n} < 1+ \frac{H_{n-1}}{n},\qquad \left(1+\frac{1}{\sqrt{n}}\right)^2=1+\frac{2\sqrt{n}+1}{n}, $$ so it is enough to show that $H_{n-1}\leq 2\sqrt{n}+1$, at least from some point on.
On the other hand the Cauchy-Schwarz inequality ensures $$ H_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}\leq\sqrt{\sum_{k=1}^{n-1}1\sum_{k=1}^{n-1}\frac{1}{k^2}} < \sqrt{n\zeta(2)} $$ and we are done.

Answer 3

This is the same as $n\le (1+1/\sqrt n)^{2n}.$ Expanding using the binomial formula, we want

$$n\le \sum_{k=0}^{2n}\binom{2n}{k}1^{2n-k}(1/\sqrt n)^k.$$

Conisder the $k=2$ term in the sum.

Answer 4

It is equivalent to prove $$ n < \left(1 + \frac{1}{\sqrt{n}}\right)^{2n}.$$

By the binomial identity, the right hand side is at least \begin{align*} & 1 + \binom{2n}{1}\frac{1}{\sqrt{n}} + \binom{2n}{2}\frac{1}{n} \\ = & 1 + 2\sqrt{n} + 2n - 1 \\ = & 2n + 2\sqrt{n} \\ > & n. \end{align*}

Answer 5

$(1+x)^n \ge (n^2/4) x^2 $, $n \ge 1$, $x \ge 0$.

Proof:

$(1+x)^n=$

$1+nx + (n(n-1)/2!)x^2+...\gt (n(n-1)/2)x^2 \ge (n^2/4)x^2,$

since we have

$(n)((n-1)/2) \ge n( n/4)$, for $n\ge 2.$

Let $x=2/√n$, in

$(1+x)^n > (n^2/4)x^2$;

$1+2/√n \gt \sqrt[n]{n}$, $n \ge 1$.

Finally

$(1+2/√n)^2 >(1+2/√n) \gt \sqrt[n]{n}$, $n \ge 1$.

Answer 6

$$\begin{align} \left(1+{1\over\sqrt n}\right)^{2n} &=\left(1+{2\over\sqrt n}+{1\over n}\right)^n\\ &\gt\left(1+{2\over\sqrt n} \right)^n\\ &\ge1+{n\choose1}{2\over\sqrt n}+{n\choose2}\left(2\over\sqrt n\right)^2\\ &\gt1+0+2(n-1)\\ &=2n-1\\ &\ge n \end{align}$$

Answer 7

Well for natural (positive) $n$ then $\sqrt[n]{n} \le (1+ \frac 1{\sqrt n})^2 \iff$

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

So $(1+\frac 1{\sqrt n})^{2n} \ge 1 + \frac {2n}{\sqrt n}$ and dang... that's not enough.

But lets go one more term.

Remember the reason $(1 + b)^n \ge 1+ nb$ is becase $(1 + b)^n = 1 + nb + C_2b^2 + C_3b^3 + ..... + b^n$ and $C_kb^k \ge 0$. (Note if $b$ is large and $n>0$, in particular if $b \ge 1$, this is $(1+b)^n \ge 1 + nb$ is a very large discrepancy in the inequality).

So $(1+\frac 1{\sqrt n})^{2n} \ge 1 + \frac {2n}{\sqrt n} + {2n \choose 2}\frac 1{n}$.

So ${2n \choose 2} = \frac {2n(2n-1)}2 = n(2n-1)$ so

$(1 + \frac 1{\sqrt n})^{2n} \ge 1 +\frac{2n}{\sqrt n} + n(2n-1)\frac 1n =$

$1 + 2\sqrt n + 2n- 1 = 2(\sqrt n + n) > n$.

.... which if we did that correctly is not even close.