Cauchy's criterion for $\sum_{n=1}^{\infty}\frac{1}{n^2}$?

Question

Cauchy's criterion for series convergence is that for a $\sum_{n=1}^{\infty} {a_n}$ if

$$\limsup\sqrt[n]{a_n} < 1$$

then the series converges. If its $>1$ then it diverges and if its $1$ then this criterion doesnt give an answer

If we have

$$\sum_{n=1}^{\infty}\frac{1}{n^2}$$

then how come that this criterion gives us an answer?

Because $\lim\sqrt[n]{a} = 1$ if $a>0$

But textbook says that the above series converges. So maybe the above limit is actually less than $1$? Or have i misinterpreted Cauchy's criterion?

Answer 1

I don't know why is it that you mentioned that $\limsup\sqrt[n]a=1$ if $a>0$. That's not relevant here. What matters is that$$\limsup\sqrt[n]{\frac1{n^2}}=\limsup_n\frac1{\sqrt[n]n^2}=1.$$

But, yes, Cauchy's criterion does not give an answer in that case. That does not mean that there is no answer; it only means that the answer must be determined by some other method. In the case of the series $\sum_{n=1}^\infty\frac1{n^2}$ you can use, for instance, the integral test. Or ot use the fact that$$(\forall n\in\Bbb N\setminus\{1\}):\frac1{n^2}\leqslant\frac1{n(n-1)}=\frac1{n-1}-\frac1n.$$

Answer 2

In this case we have

$$\lim_{n\to \infty} \left(\frac1{n^2}\right)^\frac1n=\lim_{n\to \infty}\left(\frac1{n^\frac1n}\right)^2=1$$

and the limit doesn't approach strictly from above, Cauchy's root criterion is not conclusive and we need to refer to other methods to conclude for convergence, that is for example integral test or any proof for the Basel problem

$$\sum_{n=1}^{\infty}\frac{1}{n^2} =\frac {\pi^2} 6$$

Note also that the limit for $a>0$

$$\lim_{n\to \infty} \left(a\right)^\frac1n=1$$

is not relevant here indeed we are using that

$$\lim_{n\to \infty} \left(n\right)^\frac1n=\lim_{n\to \infty} e^{\frac{\ln n}n}=e^0=1$$