If sequence converges, then terms converge to zero? I've only seen this theorem for series:

If series converges, then the terms converge to 0.

However, I've seen it applied in sequences (e.g. [1]), so does this same apply to sequences? I cannot seem to find a proof or the theorem for that matter.
 A: The point in the example you quoted is that they are considering a limit in the form

$$\lim_{n\to\infty} g(n)\sqrt n$$

Since $\sqrt n\to\infty$, in order for $g(n)\sqrt n$ to converge to $L\in\Bbb R$ you need $g(n)\to0$, because $$g(n)=\frac{g(n)\sqrt n}{\sqrt n}\to \left[\frac{L}{\infty}\right]=0$$
But it's a special instance which is  not true for a general sequence $a_n$, of course!
A: A sequence is an indexed and ordered list of terms. When it is said that an infinite sequence converges, this means the sequence reaches a finite limit (which can be zero or nonzero), i.e. $\displaystyle |\lim_{k \to \infty} a_k| < \infty$.
A series is the sum of the individual terms of a sequence. When it is said that an infinite series converges, this means the limit of the partial sums is finite (again it can be zero or nonzero) , i.e. $\displaystyle |\lim_{n \to \infty} \sum_{k=1}^{\infty} a_k| < \infty$.
In the latter case, it is necessary but not sufficient that the sequence being summed has a limit of zero - an example of this occurs with the harmonic series. Apart from the the sequence of terms having a limit of zero, other criteria need to be satisfied - this is the basis of the tests of series convergence.
An example of a sequence that converges to a nonzero limit is $\displaystyle a_1 = 1, a_{k+1}= 1+\frac{1}{a_k}$, which converges to the golden ratio, $\phi$.
