How do you show that $\sum_{n=1}^{\infty}(-1)^{n}\dfrac{3n-1}{n^2 + n} = \log\left({32}\right) - 4$?

Question

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?

Answer 1

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

Answer 2

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

• $\sum_{n=1}^\infty \frac{(-1)^{n}}n=\ln 2$

Answer 3

$$\dfrac{3n-1}{n(n+1)}=\dfrac{3(n+1)-4}{n(n+1)}=\dfrac3n-4\left(\dfrac1n-\dfrac1{n+1}\right)=\dfrac4{n+1}-\dfrac1n$$

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 $\ln(1+x)$?