# Study the converge $\sum_{n=1}^{\infty}(\sqrt[n]{n}-1)$

I need to study the convergence of the series

$$\sum_{n=1}^{\infty}(\sqrt[n]{n}-1)$$

Now, I know that if we have a series $$\sum_{n=1}^{\infty}a_n$$ with positives elements and we can find a series $$\sum_{n=1}^{\infty}b_n$$ so that $$0 then if $$\sum_{n=1}^{\infty}b_n$$ is convergent then $$\sum_{n=1}^{\infty}a_n$$ is convergent.

Else, if $$\sum_{n=1}^{\infty}a_n$$ is divergent then $$\sum_{n=1}^{\infty}b_n$$ is divergent. The problem is I do not really know how to choose the series. Can you help me out?

• Actually, $\sum_{n=1}^\infty\sqrt[n]n$ diverges. – José Carlos Santos Oct 26 '18 at 16:54
• $$\lim_{n\rightarrow \infty}n^{1/n}=1$$ so your comparison sum is divergent... – Eleven-Eleven Oct 26 '18 at 16:54

The series diverges by comparison with $$\sum \frac{1}{n}\log n$$.

Let $$n \ge 2$$. Since $$\sqrt[n]{n} = e^{\frac{1}{n}\log n}$$, the mean value theorem gives $$\sqrt[n]{n} - 1 = e^{c_n}\cdot \frac{1}{n}\log n$$ for some $$c_n\in \left(0, \frac{1}{n}\log n\right)$$. Now $$e^{c_n} > 1$$, so that $$\sqrt[n]{n} - 1 > \frac{1}{n}\log n$$ Now you can compare your series with the divergent series $$\sum \frac{1}{n}\log n$$.

• Can you detaliate how you got that $\sqrt[n]{n} - 1 = e^{c_n}\cdot \frac{1}{n}\log n$? For what function you appy mean value theorem? – Ghost Oct 26 '18 at 17:14
• @Ghost I apply it to the function $f(x) = e^x$. – kobe Oct 26 '18 at 17:15
• So for $f(x)=e^x$ we apply it on the interval $(0, \frac{1}{n}log n)$? – Ghost Oct 26 '18 at 17:18
• @Ghost correct. – kobe Oct 26 '18 at 17:20
• Now the only problem is how to show that the new series is divergent. – Ghost Oct 26 '18 at 17:23

$$\sqrt[n]n-1={n-1\over(\sqrt[n]n)^{n-1}+(\sqrt[n]n)^{n-2}+\cdots+\sqrt[n]n+1}\ge{n-1\over n+n+\cdots+n+n}={n-1\over n^2}\ge{1\over2n}$$

(with the final inequality assuming $$n\gt1$$).

• This is actually alot faster indeed. Nice thinking! But, be careful cause for $n\leq2$ the last inequality is not right. – Ghost Oct 26 '18 at 17:31
• (+1) for using elementary analysis only. – Mark Viola Oct 26 '18 at 18:25
• @Ghost, thanks for your comment. Note, however, that the final inequality is correct for $n=2$. (My parenthetical remark accounts for the one case, $n=1$, where it's not correct.) – Barry Cipra Oct 26 '18 at 19:06
• Yes, you are right! Sorry, I typed it wrong but, for $n<2$ it is not right. Now if we take n as a natural number, the only wrong case is $n=1$ but for n real it is $n<2$. Sorry I did not take into consideration n is natural number. Anyway, great thinking! – Ghost Oct 26 '18 at 20:22

Hint: $$n^{1/n}-1=\left(e^{\log n/n}-1\right)\sim\frac{\log n}{n}$$