Does a series converge if its initial value is undefined? I understand that the series $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.
What happens when you start at a different value than n=1?
For example, does the series $\sum_{n=0}^\infty \frac{1}{n^2}$ still converge even though it starts with a value that is undefined?
My highest level of math is Calculus II, and I'm new to this site and don't quite understand a lot of the notation that people use to answer math questions, so bare with me please!  Thanks for the help in advance!
 A: The expression $\sum_{n=0}^\infty\frac1{n^2}$ does not make sense because the first term involves a division by zero and it is undefined.
Nevertheless, if you start the series from any positive integer, it also converges.
In fact, it is not difficult to show that, given a series $\sum_{n=1}^\infty a_n$ and any positive integer $k$, the series $\sum_{n=1}^\infty a_n$ converges if and only if the series $\sum_{n=k}^\infty a_n$ converges.
A: Convergence/divergence of an infinite series $\sum_{n=n_0}^{\infty} a_n$ is determined — by definition! — by its behavior as $\color{red}{n\to\infty}$. (More precisely, by the limit of its partial sums.) What you're asking has nothing to do with convergence or divergence. You've simply set up an expression that doesn't make sense because one of its terms doesn't make sense: a numerical series must consists of numbers, and $1/0$ is NOT a number. So sorry, but the question "What happens if we start with $n=0$" is moot in this case.
It's just like discussing the value of something like $\displaystyle \frac{1}{0}+\sqrt{16}$. Since the first term is undefined, the calculation stops right there, and we don't need even to think about what the symbols $+$ or $\sqrt{\phantom{16}}$ represent.
That being said, the question "What happens if we start with other values of $n_0$?" is a very good one! And it's already addressed and answered in the response from @ajotatxe.
