I know
$
\sum_{n=2}^\infty \frac{1}{n*ln(n)}
$
is divergent by the integral test or comparison test; however, I notice that it fails the Series Test For Divergence ($\lim_{n\to\infty}a_n \neq 0 \Rightarrow Divergence$). Can a series fail this test and still diverge?
[Math] A Series Fails The Test For Divergence, but is Still Divergent
divergent-seriessequences-and-series
Related Solutions
What you are noting is that the series $$\sum_{i=0}^\infty \,\frac{(-1)^n}{n}\quad {\bf {converges},}$$ as you found by the alternating series test, but does not converge absolutely: $$\sum_{i=0}^\infty \,\left|\frac{(-1)^n}{n}\right| \quad = \quad \sum_{i = 0}^\infty\,\frac 1n\quad\bf{does\; not\; converge.}$$
Note: the $p$-series test is applicable for sums of the form: $\displaystyle\sum \frac 1{n^{p}},$ and your "given" series does not "fit" that form for odd $n$; indeed, the most appropriate test to use here, as you used in the end, is the alternating series test.
The proof in the paper ``Neither a worst convergent series nor a best divergent series exists" by J. Marshall Ash that I referenced above is so nice that I wanted to reproduce it below before it gets lost on the Internet.
$\bf{Theorem: }$ Let $\sum_{n=1}^{\infty} c_n$ be any convergent series with positive terms. Then, there exists a convergent series $\sum_{n=1}^{\infty} C_n$ with much bigger terms in the sense that $\lim_{n\rightarrow\infty} C_n/c_n = \infty$. Similarly, for any divergent series $\sum_{n=1}^{\infty} D_n$ with positive terms, there exists a divergent series $\sum_{n=2}^{\infty} d_n$ with much smaller terms in the sense that $\lim_{n\rightarrow\infty} \frac{d_n}{D_n} = 0$.
$\bf{Proof: }$ For each $n$, let $r_n = c_n + c_{n+1}+\cdots$ and $s_n = D_1 + \cdots + D_n$. Letting $C_n = \frac{c_n}{\sqrt{r_n}}$ and $d_n = \frac{D_n}{s_{n-1}}$, then $\lim_{n\rightarrow\infty} \frac{C_n}{c_n} = \lim_{n\rightarrow\infty} \frac{1}{\sqrt{r_n}}=\infty$ and $\lim_{n\rightarrow\infty} \frac{d_n}{D_n} = \lim_{n\rightarrow\infty} \frac{1}{s_{n-1}} = 0$, so it only remains to check $\sum C_n$ converges and that $\sum d_n$ diverges. To see that this is indeed the case, simply write $C_n = (1/\sqrt{r_n})(r_n-r_{n+1})$ and $d_n = 1/s_{n-1}(s_n-s_{n-1})$; observe that $\int_0^{r_1} 1/\sqrt{x}dx<\infty$ and $\int_{s_1}^{\infty} 1/xdx = \infty$; and note that the $n$th term of series $\sum C_n$ is the area of the gray rectangle in Figure 1a, while the $n$th term of series $\sum d_n$ is the area of the gray rectangle in Figure 1b.
Best Answer
Yes, decay of the summand as $ n \rightarrow \infty$ is a necessary, not sufficient condition for convergence of the series.