Understanding Divergence Test through difference of consecutive partial sums

Question

It's known that the difference between two consecutive partial sums $s_{n+1}$ and $s_n$ should give us the sequence term $a_{n+1}$, i.e. $s_{n+1} - s_n = a_{n+1}$.

If we take a limit of both sides as $n$ approaches infinity, we get:

\begin{equation} \lim_{n \to \infty} s_{n+1} - s_n = \lim_{n \to \infty} a_{n+1}. \end{equation}

Let's say it's also known that $\lim_{n \to \infty} a_n = 0$. That would imply that $\lim_{n \to \infty} a_{n+1} = 0$, and $\lim_{n \to \infty} s_{n+1} - s_n = 0$.

This seems to say that $s_{n+1} \approx s_n$ for some large value of $n$ and the sequence of partial sums (infinite sum of $a_n$) converges to some value. Two questions:

1. Why would the Divergence Test be inconclusive if it seems like the limit of partial sums converges to some value? (I understand that the harmonic series diverges although the limit of $a_n$ goes to zero, so looking for an argument along the lines of partial sums.)

2. Why is the same limit of $a_n$ going to zero enough to say that an alternating series converges? (i.e. then we have $\lim_{n \to \infty} s_{n+1} - s_n = \lim_{n \to \infty} (-1)^n a_{n+1}$ and the same argument applies).

Answer 1

As the harmonic series illustrates, a sequence's general term can go to zero without its being Cauchy. It doesn't seem like the partial sums converge, only that the difference of successive partial sums goes to zero.

In the case of an alternating decreasing series, it turns out the series will be Cauchy. See the (clever) proof of the Leibniz test. I would reproduce it, but it should be easy to find. I learned it in Best and Penner's Calculus, a fine book.