<s>[INST] How do you show that $\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$? how to show that $$\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$$? Can I use the Alternating Series test and how? 
 [/INST] $$S_1= \sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n}$$
$$\dfrac{3n-1}{n^2 + n} = \dfrac{3-\frac{1}{n}}{n + 1} = \left(\frac{4}{n+1} - \frac{1}{n}\right)$$
So
\begin{align}
S_1 &= \sum_{n=1}^{\infty}(-1)^{n}\left(\frac{4}{n+1} - \frac{1}{n}\right)\\
 &= 4\underbrace{\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n+1}}_{s_2} - \underbrace{\sum_{n=1}^{\infty}(-1)^{n}\frac{1}{n}}_{-\ln(2)}
\end{align}
We know, then
$$\ln(1 + x) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^k}{k} = x - \frac{x^2}{2} + \frac{x^2}{3} - \frac{x^2}{4} + \cdots$$
and for $x = 1$ we have
$$\ln(1 + 1) = \ln(2) = \sum_{k=1}^{\infty}(-1)^{k+1}\frac{1^k}{k} = 1 \underbrace{- \frac{1^2}{2} + \frac{1^2}{3} - \frac{1^2}{4} + \cdots}_{s_2} = 1- s_2$$
Which implies
$$s_2 = \ln(2) -1$$
Therefore
\begin{align}
S_1 &= 4(s_2) - (-\ln(2))\\
&= 4(\ln(2) -1) - (-\ln(2))\\
&= 4\ln(2) -4 + \ln(2)\\
&= 5\ln(2) -4\\
&= \ln(32) -4\\
\end{align}

See:
the sum: $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$ using Riemann Integral and other methods
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