if terms of an alternating series approach 0 then that alternating series converges. Is it sufficient that for any alternating series to converge then limit of the its terms approach 0? If no then what is counter example of an alternating  series with limit of terms approaching 0 and not converging?
 A: To ensure convergence you can ask the extra condition that the sequence be monotone decreasing, which is sufficient but is not  necessary $^{(1)}$. For a counterexample, take $$a_{2n}=\frac{1}{2^n}$$ $$a_{2n+1}=\frac 1 n$$
Of course I mean you want to look at $$\sum (-1)^n a_n$$

$(1)$ To see this take $a_{2n}=2^{-n},a_{2n+1}=3^{-n}$.
A: No. A series $\sum_{n=1}^\infty (-1)^n a_n$ with $a_n \geq 0$ for all $n$ and $\lim_{n\to \infty}a_n$ may still diverge. Consider $\sum_{n=2}^\infty (-1)^n a_n$ with $a_{2n} = 1/n$ and $a_{2n+1} = 1/2n$ for all $n$. Then the $2n+1$-st partial sum looks like
\begin{align*}
S_{2n+1} &= (1/1 - 1/2) + (1/2 - 1/4) + (1/3 - 1/6) + (1/4 - 1/8) + \ldots + (1/n - 1/(2n)) \\
&= 1/2 + 1/4 + 1/6 + \ldots + 1/2n = (1/2)\sum_{k=1}^n 1/k
\end{align*}
Since $\lim_{n\to\infty} S_{2n+1} = (1/2)\sum_{k=1}^\infty 1/k = \infty$, the series diverges.
A second condition that ensures convergence (and is included in the Alternating Series Test) is that the sequence $\{a_n\}$ is decreasing. If you look at a proof of the AST, the reason why this works is that it ensures the $N$-th remainder $\sum_{n=N+1}^\infty (-1)^n a_n$ can be approximated by $a_n$.
