If a series is infinite at some early point can it converge? Suppose I have a series defined :
$\sum_{n=1}^{\infty}100/ln(n)$.
My intuition tells me it diverges because at n equals 1 the term is infinite (according to the partial sum definition), however my teacher taught me this:
If N is a positive integer, then the series
$\sum_{n=1}^{\infty}a_n$ 
and 
$\sum_{n=N+1}^{\infty}a_n$
both converge or both diverge.
Which one is correct?
 A: 
Suppose I have a series defined :
  $\sum_{n=1}^{\infty}100/ln(n)$

This doesn't make sense since $\ln 1 = 0$ so $\tfrac{100}{\ln n}$ isn't defined at $n=1$.

if N is a positive integer, then the series
  $\sum_{n=1}^{\infty}an$ 
  and 
  $\sum_{n=N+1}^{\infty}an$
  both converge or both diverge.

Try to understand what this means: convergence is only determined by "the tail" of the series, not by a finite number of terms in the beginning since the sum of a finite number of numbers, is finite as well.
A: Your mistake is on the very first line.
You have not defined a series.
In a legitimate series every term is a real number — or complex or in some other fixed space.
At $n=1$ you don't have a well defined number.
In a series every term is finite.
A: For the first series, you should write
$$\sum_{n\ge 2}\frac{100}{\ln(n)}$$
Yes, If the series are well defined,
they will have the same nature since $$\sum_{n=1}^ma_n=\sum_{n=1}^Na_n+\sum_{n=N+1}^ma_n$$
When $m\to+\infty$, both sides will converge or both diverge.
$\sum_{n=1}^Na_n$ do not depend on $m$.
