Limit of infinite series So if an infinite series converges to any real than the limit of the partial sums sequence is equal to $0$, right?
Does that mean that the limit of the series is $0$ too?
What confuses me is "is the limit of an infinite series equal to it's sum?"
so if the limit of the partial sums sequence is $0$, then the sum of the sequence is $0$, which allows me to say that the limit of the series is $0$, so the series converges to $0$?
 A: By definition, the sum of an infinite series is the limit of its partial sums:
$$\sum_{i = 0}^{
\infty}a_i := \lim_{N \to \infty}\sum_{i = 0}^Na_i$$
Now, in order for this limit to converge to a real number $L$, it would certainly have to be that the distance between each partial sum and the next gets arbitrarily small - after all, that's a necessary condition for any limit to converge.
$$\sum_{i = 0}^{N + 1}a_i - \sum_{i = 0}^Na_i = a_{N + 1}$$
So the values of $a_N$ must go to zero. That means that it is not correct to say that the limit of the partial sums must be zero, but it is correct to say that the limit of the terms of the series must be zero.
There is a point of confusion here, though. You keep referring to the limit of a series. That's not typical terminology, because it isn't clear whether you mean the limit of the terms (because you're not mentioning adding things up) or the limit of the partial sums (because you're calling it a "series" and you usually sum series). Generally, we say limit of a sequence and sum of a series. So, to put things as precisely as I can:
Let $a_0, a_1, \ldots, a_i, \ldots$ be an infinite sequence. The limit of the sequence is $\lim_{i \to \infty}a_i$. The sum of the series is $\sum_{i = 0}^{\infty}a_i = \lim_{N \to \infty}\sum_{i = 0}^Na_i$. If the sum of the series converges to a real number, then the limit of the sequence must be zero. Because the sum of the series may be nonzero, the limit of the sequence is not necessarily the same as the sum of the series.
A: What you are saying is not correct. If infinite series converged then partial sum converged to sum. You can say remainder terms goes to zero. Or nth terms tends to zero for large n.
A: No - the limit of the sequence of partial sums does not have to be $0$. 
For example $$\sum\limits_{i=1}^n \frac1{2^i} = \frac{2^n-1}{2^n} = 1 - \frac{1}{2^n}$$ has the partial sums being $\frac12,\frac34,\frac78, \cdots$, with an obvious limit of that sequence and of the series being $1$
It is true that for the series to converge to a finite limit then the individual terms of the series must converge to $0$, though not necessarily the other way round
A: What you are saying is not correct, recall indeed that by definiton we have
$$\sum_{k=0}^\infty a_k = \lim _{n\to \infty} S_n= \lim _{n\to \infty} \sum_{k=0}^n a_k$$
and


*

*$\sum_{k=0}^\infty a_k$ is the (infinite) series

*$S_n= \sum_{k=0}^n a_k$ is the partial sum


when the series converges, that is $$\lim _{n\to \infty} S_n=L\in \mathbb{R}$$ we can derive the necessary condition for convergence 
$$a_n=S_n-S_{n-1} \to L-L=0$$
