Cauchy condensation test - do I have to prove $\sum_{n=2}^{\infty} \frac{1}{n^2}$ converges? My original question:
$$ \sum_{n=2}^\infty \frac {1} {n (\ln(n))^2}$$
I reached after a while to this (while using Cauchy condensation test)
$$ \sum_{n=2}^\infty \frac {1} {n (\ln(n))^2} < \sum_{n=2}^{\infty} \frac {1} {n^2} $$
Now, since $\sum_{n=2}^\infty \frac{1}{n^2}$  converges, so does the original one.
But my question is, do I have to prove $\sum_{n=2}^{\infty} \frac{1}{n^2}$ converges? Or it is a sure fact? For all my exercises I didn't do it (neither did the practitioner), but just to make sure.
Is there any need? Or it is a sure fact?
 A: Due to help, I reached the solution:
I do need to prove converges of: $$ \sum_{n=2}^{\infty} \frac {1} {n^2} $$
I am doing so by using once more Cauchy condensation test:
I will call $a_n$: $$a_{k}=\frac {1} {k^2} $$
Thus, by using Cauchy condensation test($ \sum_{n=0}^{\infty}{2^k} ⋅ a_{2^k}) $, we get:
$$ a_{2^k}=\frac {1} {n^2} $$
Then use the formula of Cauchy condensation test itself, and we get:
$$ \sum_{n=2}^{\infty} \frac {1 ⋅ 2^k} {2^k ⋅ 2^k} $$
Which equals to:
$$ \sum_{n=2}^{\infty} \frac {1} {2^k} $$
And here we see it.
The $ \sum_{n=2}^{\infty} \frac {1} {2^k} $ converges.
A: Well at some point you'll have to decide what is part of your toolbox and what's still need to be proved. This will of course evolves as your own knowledge increases.
Some classical results that are also commonly accepted without proof at some point are:

*

*Geometric series $\sum c^n$ when $|c|<1$

*Geometric series with power factor $\sum n^a\,c^n$ when $|c|<1$ for any $a$

*Riemann series $\sum \frac 1{n^a}$ when $a>1$

*Bertrand series (which is a generalisation of your exercise) $\sum \frac 1{n^a\ln(n)^b\ln(\ln(n))^c\cdots}$ for $(a>1)$ or $(a=1,b>1)$ or $(a=b=1,c>1)$ or $\cdots$
Convergence for Riemann series is easily provable by comparison to an integral:
Indeed for $f\searrow$ we have
$$\int_n^{n+1}f(x)\,dx\le\int_n^{n+1}f(n)\,dx\ =\ f(n)\ = \int_{n-1}^{n}f(n)\,dx\le\int_{n-1}^{n}f(x)\,dx$$
Apply to $f(n)=\frac 1{n^a}$ and since the anti-derivative of $x^{-a}$ is in $x^{1-a}$ it is convergent only for $1-a<0\iff a>1$ at infinity.
A: $$
\sum_{n=2}^\infty \frac 1 {n^2} \le \sum_{n=2}^\infty \frac 1 {n(n-1)} = \sum_{n=2}^\infty \int_{n-1}^n \frac{dx}{x^2} = \int_1^\infty \frac{dx}{x^2} = 1 < +\infty.
$$
(Proving that $\sum_{n=1}^\infty \frac 1{n^2} = \frac {\pi^2} 6$ is a famous hard problem. It can be solved by elementary means, but only if you're very clever.)
Postscript: A variation on this theme:
$$
\sum_{n=2}^\infty \frac 1 {n^2} = \sum_{n=1}^\infty \int_{n-1}^n \frac 1 {n^2} \, dx \le \sum_{n=2}^\infty \int_{n-1}^n \frac 1 {x^2} \, dx= \cdots
$$
