Soft Question Regarding the Divergence of $\sum\limits_{k=1}^{\infty}\frac{(-1)^k k}{3k+2}$ Determine whether $\sum\limits_{k=1}^{\infty}\frac{(-1)^k k}{3k+2}$ converges or diverges.
Consider the function $f(x)=\frac{x}{3x+2}$ that generates the unsigned terms of our series.
Taking the derivative of $f$ we have $f'(x)=\frac{2}{(3x+2)^2}$
Since $f'(x)>0$ for all $x$, we know that $f(x)$ is increasing for $x>1$, therefore the alternating series test is inconclusive.
Taking $\lim\limits_{x\rightarrow\infty}f(x)=\lim\limits_{k\rightarrow\infty}\frac{k}{3k+2}=\frac{1}{3}$ we see that by the divergence test, our series must diverge.
I'm fairly confident that I've reached the correct conclusion here. However, some subsequent computing has piqued my curiosity.

Certainly the plot itself doesn't say a ton but after computing
$$S_{10^2}\approx0.08320124$$
$$S_{10^3}\approx0.08418853$$
$$S_{10^4}\approx0.0842884$$
$$S_{10^5}\approx0.0842984$$
$$S_{10^6}\approx0.0842994$$ the series certainly seems to be converging to something. Am I seeing something in the wrong way here or is it just a case of the series diverging so slowly as to appear like it is converging? I realize this kind of soft question is generally discouraged here but I was genuinely startled by this series and was hoping for some clarity.
Sincerely,
calculus student trying to build intuition.
 A: $|a_k| = \frac{k}{3k+2} \to \frac{1}{3} \neq 0$ series doesn't converge
EDIT In case you need something deeper, multiply numerator and denominator by 3 and add/subtract 2:
$$
\frac{1}{3} \sum_k \frac{(-1)^k \cdot (3k+2)}{3k+2} - \frac{2}{3} \sum_k \frac{(-1)^k}{3k+2}
$$
The second sum $\to_n a \log 2$ for some constant $a$, the first sum alternates between $\frac{1}{3}$ and $-\frac{1}{3}$.
A: Look at the sum of the first $n$ terms.
$\begin{array}\\
\sum\limits_{k=1}^{n}\frac{(-1)^k k}{3k+2}
&=\sum\limits_{k=1}^{n}(-1)^k\frac{ k}{3k+2}\\
&=\sum\limits_{k=1}^{n}(-1)^k\left(\frac{ k}{3k+2}-\frac13+\frac13\right)
\qquad\text{(because the terms tend to } \frac13)\\
&=\sum\limits_{k=1}^{n}(-1)^k\left(\frac{ -2}{3(3k+2)}+\frac13\right)\\
&=\sum\limits_{k=1}^{n}(-1)^k\left(\frac{ -2}{3(3k+2)}\right)+\sum\limits_{k=1}^{n}(-1)^k\frac13\\
&=-\frac23\sum\limits_{k=1}^{n}\frac{ (-1)^k}{3k+2}+\frac13\sum\limits_{k=1}^{n}(-1)^k\\
\end{array}
$
The first sum converges
by the alternating series test,
while the second sum
alternates between
$-\frac13$ and $0$.
Therefore the sum does not converge.
Your plots seem to show convergence
probably because you only tested
even values of $n$.
A: More complex for the same result.
Consider the partial sum
$$S_p=\sum\limits_{k=1}^{p}(-1)^k\frac{ k}{3k+2}$$ $$S_p=\left(\frac{1}{6}-\frac{2 \pi }{9 \sqrt{3}}+\frac{2 \log (2)}{9}\right)+\frac{(-1)^p}{18}  \left(3-4 \Phi \left(-1,1,p+\frac{5}{3}\right)\right)$$ where appears the Lerch transcendent function.
The first term
$$\frac{1}{6}-\frac{2 \pi }{9 \sqrt{3}}+\frac{2 \log (2)}{9}\approx -0.082367151927616120262$$
The second term
$$\frac{1}{18}  \left(3-4 \Phi \left(-1,1,p+\frac{5}{3}\right)\right)$$ is an increasing function which is asymptotic to $\frac 16$ but the problem is the $(-1)^p$.
As @Marty Cohen commented, using only even values of $p$, you would get for the limit of $S_{2p}$
$$\frac{9-2 \sqrt{3} \pi +6 \log (2)}{27}=0.084299514739050546405$$ which is what you obtained.
Repeat your calculations for $S_{10^k+1}$
If you consider the limit for odd values of $p$, you would get
$$-\frac{2}{27} \left(\sqrt{3} \pi -3 \log (2)\right)=-0.24903381859428278693$$
What is interesting is to notice that
$$S_{2p+1}-S_{2p}=-\frac 13 +\frac{2}{3(6 p+5)}$$
