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I am given this statement and I need to prove it. the statement is for all $n\ge 1$: $$\sqrt[n]{n}\leq 1+\frac{2}{\sqrt{n}}$$

I am trying to prove with induction. But I am stuck for step n=k+1, how can I then decompose the step: $\sqrt[n+1]{n+1}$?

Thanks for help in advance

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$\sqrt[1]{x}=x$. – Hanul Jeon Jan 6 '13 at 13:06
up vote 7 down vote accepted

Induction does not seem to suggest itself for this kind of task. Try to find out something about $$\left(1+\frac2{\sqrt n}\right)^n\stackrel?\ge n$$ instead (using binomials).

EDIT: To summarize below comments into this solution, there are (at least) two methods to show $\left(1+\frac2{\sqrt n}\right)^n\ge n$ for all $n\ge 1$.

One is to use binomial expansion $$(a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k}.$$ here with $a=1$, $b=\frac2{\sqrt n}$. If $n\ge2$ (noting that of course the remaining case $n=1$ can be checked directly) and we drop the nonnegative summands belonging to all $k>2$, we find $$\begin{align}\left(1+\frac2{\sqrt n}\right)^n&\ge{n\choose 0}+{n\choose 1}\frac2{\sqrt n}+{n\choose2}\frac 4n\\&=1+2\sqrt n+2(n-1)\\&=n+(n+2\sqrt n-1)\\&\ge n.\ _\square\end{align}$$

Another way is to make use of the famous and very important Bernoulli inequallity $$ (1+x)^r\ge 1+rx\quad\text{for all }x\ge-1, r\ge 0.$$ If we plug in $x=\frac2{\sqrt n}$ (of course) and $r=\frac n2$ (a bit surprisingly), we obtain $$\left(1+\frac2{\sqrt n}\right)^{\frac n2} \ge 1+\sqrt n$$ and by squaring this $$\left(1+\frac2{\sqrt n}\right)^{n} \ge (1+\sqrt n)^2=1+2\sqrt n + n> n.\ _\square$$

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thanks, now i said: $(1+\dfrac{2}{\sqrt{n}})^n>\dfrac{n^2 \cdot (\dfrac{2}{\sqrt{n}})^2}{4} = n$, thats why the inequality $(1+\dfrac 2{\sqrt n})^n \ge n$ is okay. am i right? – doniyor Jan 6 '13 at 13:24
It depends on how you came up with that inequality. I would have used $(1+x)^n= 1+nx+\frac{n(n-1)}2x^2+\ldots\ge 1+nx+\frac{n(n-1)}2x^2$ for $x\ge 0$. Or you could use Bernoulli's inequality $(1+x)^r\ge 1+rx$ (if $r\ge0, x\ge -1$) with $x=\frac2{\sqrt n}$ and $r=\frac n2$. Or probably a lot more. – Hagen von Eitzen Jan 6 '13 at 13:34
@doniyor I agree with Hagen, using Bernoulli's inequality is extremely useful when dealing with nth powers (or roots), especially when it has the form $1+x$. Even if it doesn't, you can manipulate it to that. For example, if it has form $a+x$, divide by $a$. If it has form $x$, split it into $1+(x-1)$. As a side note, the resin why Hagen used $r= \frac {n}{2}$ is because this inequality is tighter for small numbers, and using $r=n$ will be too loose of a bound. – Calvin Lin Jan 6 '13 at 16:11
@Calvin, yeah you are right. thanks – doniyor Jan 6 '13 at 17:11

I think Hagen's answer is better than this one, but this one illustrates a useful general principle: take logarithms to simplify exponentials (and products). A second useful principle is to use power series to get inequalities for complicated functions in terms of simple polynomials.

In this case, since $\ln$ is an increasing function, the inequality $\sqrt[n]n \le 1 + 2/\sqrt n$ is equivalent to $\ln(\sqrt[n]n) \le \ln(1 + 2/\sqrt n)$, or $\frac1n \ln n \le \ln(1 + 2/\sqrt n)$. Now we use the inequality $\ln(1+x) \ge x - x^2/2$, valid for all $0\le x\le1$; one way to see this is by noting that the power series $\ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + \cdots$ is an alternating series of decreasing terms in that range. Thus it suffices to show that $$ \frac{\ln n}n \le \frac2{\sqrt n} - \frac12\bigg( \frac2{\sqrt n} \bigg)^2, $$ or equivalently that $$ \ln n + 2 \le 2\sqrt n; $$ this is much easier to check than the original problem.

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