# Convergence of the series $\sqrt[n]n-1$

Let $a_n=\sqrt[n]n-1$. Does $\sum_{n=1}^\infty a_n$ converge?

## 6 Answers

Hint: $$\sqrt[n]n = e^{\frac{\log n}{n}} > 1+\frac{\log n}{n}$$

• How was this taken ? Commented Jun 12, 2013 at 16:05
• Not sure what you mean, @RajindaWickrama. Commented Jun 12, 2013 at 16:16
• How is e^(ln(n)/n) > 1 + ln(n)/n Commented Jun 12, 2013 at 16:23
• Since $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$, when $x>0$, this means that $e^x > 1+x$, since $1+x$ is just the first two terms of the series. Then $x=\frac{\log n}{n}$ to get our result. Commented Jun 12, 2013 at 16:25
• Thanks for the explanation Commented Jun 12, 2013 at 16:33

$$a_n=n^{1/n}-1= e^{\log{n}/n}-1 \sim \frac{\log{n}}{n}$$

The sum diverges by comparison with the sum of $b_n=1/n$.

From $\left(1+\frac1n\right)^n\to e$, we conclude $$\left(1+\frac1n\right)^n<3<n=(1+a_n)^n$$ for almost all $n$. Hence $a_n>\frac1n$ for almost all $n$ and $\sum a_n$ diverges.

Remark: We don't even need the introduction of $e$. The observation $$\left(1+\frac1n\right)^n=\sum_{k=0}^n{n\choose k}n^{-k}\le \sum_{k=0}^n\frac1{k!}<1+\sum_{k=1}^ \infty 2^{1-k} =3$$ suffices for an "elementary" approach.

• The "observation" is very nice, I don't think I'd seen that. Commented Jul 3, 2013 at 6:11

$$\sqrt[n]{n}-1=\frac{n-1}{\sqrt[n]{n^{n-1}}+\sqrt[n]{n^{n-2}}+...+1}\geq \frac{n-1}{n\sqrt[n]{n^{n-1}}}=\frac{n-1}{n}\frac{1}{n^{\frac{n-1}{n}}}=\frac{n-1}{n}\frac{\sqrt[n]{n}}{n}>\frac{n-1}{n}\frac{1}{n}$$

Now use that for all $n >2$ we have

$$\frac{n-1}{n}\frac{1}{n}>\frac{1}{2}\frac{1}{n} \,,$$

or limit compare it to the harmonic series.

• @ThomasAndrews Fixed, ty. Commented Jun 10, 2013 at 18:09

For $n\ge2$, \begin{align} n(\sqrt[n]{n}-1) &\ge n(2^{1/n}-1)\\ &\to\log(2) \end{align} Thus, for some $N$, if $n\ge N$, then $n(\sqrt[n]{n}-1)\ge\frac12$. Therefore, for $n\ge N$, $$\sqrt[n]{n}-1\ge\frac1{2n}$$ Thus, the series diverges by comparison to the harmonic series.

Note that for any $n\ge3$: $$n>e$$ and so, $$n^{1/n} > e^{1/n}$$

And since we know: $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ we can say: $$e^{1/n} = \sum_{k=0}^\infty \frac{1}{k!\;n^k} = 1 + \frac{1}{n} + \frac{1}{2n^2}+\cdots$$ Which, for $n\in\Bbb N$, is surely larger than the truncated series $1+1/n$, and so follows the strict inequality: $$n^{1/n} > e^{1/n} > 1 + \frac{1}{n}$$

Subtract one from both sides, yielding: $$\sqrt[n]{n}-1 = n^{1/n} - 1>\frac1n$$

Which is true $\forall n\ge 3$. Evaluate the sum: $$\sum_{n\ge 3} (\sqrt[n]{n}-1) > \sum_{n\ge 3} \frac{1}{n}$$

Thus, $\sum a_n$ is larger than the divergent harmonic sum, and is divergent too.