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? 
 A: $$\dfrac{3n-1}{n(n+1)}=\dfrac{3(n+1)-4}{n(n+1)}=\dfrac3n-4\left(\dfrac1n-\dfrac1{n+1}\right)=\dfrac4{n+1}-\dfrac1n$$
This can also be achieved by Partial Fraction Decomposition $$\dfrac{3n-1}{n(n+1)}=\dfrac An+\dfrac B{n+1}$$
Now $\ln(1-x)=-\sum_{r=1}^\infty\dfrac{x^r}r$ for $-1\le x<1$  by 
What is the correct radius of convergence for $\ln(1+x)$?
A: HINT
We have that
$$(-1)^{n}\dfrac{3n-1}{n^2 + n} =(-1)^{n}\frac1n\dfrac{3n+3-4}{n + 1} =3\frac{(-1)^{n}}n-4(-1)^{n}\dfrac{1}{n(n + 1)}=\ldots$$
and since by telescoping
$$\dfrac{1}{n(n + 1)}=\frac1{n}-\frac1{n+1}$$
we obtain
$$-\frac{(-1)^{n}}n+4\dfrac{(-1)^{-1}}{n + 1}$$
then recall that by alternating harmonic series


*

*$\sum_{n=1}^\infty \frac{(-1)^{n}}n=\ln 2$
A: $$S_1= \sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n}$$
$$\dfrac{3n-1}{n^2 + n} = \dfrac{3-\frac{1}{n}}{n + 1} = \left(\frac{4}{n+1} - \frac{1}{n}\right)$$
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
