The convergence of the series $\frac{1}{n(\ln(n))^2}$ depends on the starting index? For the series
$$\frac{1}{n(\ln(n))^2}$$
If I want to check the convergence of the series using integral test :
Can I start the integration from 1 ( then the series diverges ) ? or this is not allowed since the function tends to infinity at x=1?
$$\int_{1}^{\infty}\frac{1}{x(\ln(x))^2}dx$$
Do I must start from x=2 ? ( the series converges)
Note: it is not mentioned in the problem the beginning of the summation so I am confused ..
 A: The series $$\sum_{n_0}^{\infty}\frac{1}{n(\ln(n))^2}$$
is well defined only starting from $n=2$ (since $\log(1)=0$).
Whereas the corresponding improper integral is well defined also starting from 1 (the reason is that improper integrals are defined as limits):
$$\int_{1}^{\infty}\frac{1}{x(\ln(x))^2}dx$$
Note also that the convergence of a series is not affected by a finite numbers of initial terms, thus the series:
$$\sum_{2}^{\infty}\frac{1}{n(\ln(n))^2}$$
$$\sum_{1000}^{\infty}\frac{1}{n(\ln(n))^2}$$
have the same behaviour.
A: the integral test is the following:
consider the function $f$, if $f$ is continuous non-negative function that is defined at $[N,\infty)$ and is monotone decreasing at this interval then:
the series $\sum\limits_{x=N}^\infty f(x)$ converge if and only if $\int\limits_N^\infty f(x) dx$ is finite.
from this we can understand that no, the starting point is irrelevant as long as $f(N)$ exists, in your case $f(1)$ doesn't exists hence we can't start with it.
the reason that it is irrelevant is this:
consider $\int\limits_{N_1}^\infty f(x) dx=L$(where $L$ is a real number), this implies that $\sum\limits_{x=N_1}^\infty f(x)=K$(where $K$ is a real number)
now consider the integer $N_2$, if we assume that $\forall x\in[N_2,\infty)f(x)$-exists, then when $N_2$ is greater than $N_1$ it is obvious that $\int\limits_{N_2}^\infty f(x) dx<L$, because $f$ is monotone decreasing.
if $N_2<N_1$ then  $\int\limits_{N_2}^\infty f(x) dx=\int\limits_{N_2}^{N_1} f(x) dx+\int\limits_{N_1}^{\infty} f(x) dx=\int\limits_{N_2}^{N_1} f(x) dx+L$ and because the function is defined at $x\in[N_2,N_1]$ we know that $\int\limits_{N_2}^{N_1} f(x) dx$ also exists, hence $\int\limits_{N_2}^{N_1} f(x) dx+L$ exists.
so we can start the integration at any arbitrary integer as long as it the function exists at that point, and if the function exists between 2 points(including the points) and you know that the integral is finite when starting from one of the points you can know that it exists also when starting from the second point
