Calculate the next sum $\sum\limits_{k=3}^\infty (-1)^k \frac{3k+1}{k(k+1)}(x-1)^k$

Question

Find the next sum $$\sum_{k=3}^\infty (-1)^k \frac{3k+1}{k(k+1)}(x-1)^k$$

I do not know how to solve this. I’ve tried to split the fraction into two fractions and use $\ln(x)=\int_1^x dt/t$, but I’m stuck here. So any suggestion will be appreciated.

Answer 1

Hint

$\sum_{k=3}^\infty (-1)^k \frac{3k+1}{k(k+1)}(x-1)^k$

$\sum_{k=3}^\infty (-1)^k \frac{2k + (k+1)}{k(k+1)}(x-1)^k$

$\sum_{k=3}^\infty (-1)^k \frac{1}{k}(x-1)^k + \frac 2 {(x-1)}\sum_{k=3}^\infty (-1)^k \frac {(x-1)^{k+1}}{(k+1)}$

$-\sum_{k=3}^\infty (-1)^{k+1} \frac{(x-1)^k}{k}+ \frac 2 {(x-1)}\sum_{k=3}^\infty (-1)^{k+2} \frac {(x-1)^{k+1}}{(k+1)}$

$-\sum_{k=3}^\infty (-1)^{k+1} \frac{(x-1)^k}{k}+ \frac 2 {(x-1)}\sum_{k=4}^\infty (-1)^{k+1} \frac {(x-1)^{k}}{(k)}$

$- \frac {(x-1)^3}{3}-\sum_{k=4}^\infty (-1)^{k+1} \frac{(x-1)^k}{k}+ \frac 2 {(x-1)}\sum_{k=4}^\infty (-1)^{k+1} \frac {(x-1)^{k}}{(k)}$

$- \frac {(x-1)^3}{3}+ (-1+ \frac 2 {(x-1)})\sum_{k=4}^\infty (-1)^{k+1} \frac {(x-1)^{k}}{(k)}$

Then take a look at the developpement of ln (x) for the serie...

Answer 2

\begin{align} \dfrac{1}{1-y} &= \sum_{k=0}^\infty y^k \hspace{3cm}\text{|y|<1}\\ \dfrac{1}{(1-y)^2} &= \sum_{k=1}^\infty ky^{k-1} \hspace{3cm}\text{|y|<1}\\ \dfrac{3y}{(1-y)^2}+\dfrac{1}{1-y} &= \sum_{k=0}^\infty (3k+1)y^k \\ \dfrac{2y+1}{(1-y)^2}&= \sum_{k=0}^\infty (3k+1)y^k \\ \dfrac{4-y}{(1-y)^2} &= \sum_{k=1}^\infty (3k+1)y^{k-1} \hspace{2cm}\text{here take integration}\\ \dfrac{3y}{1-y}-\ln(1-y) &= \sum_{k=1}^\infty \dfrac{3k+1}{k}y^k \hspace{3cm}\text{0<y<1}\\ \dfrac{3y}{1-y}-\ln(1-y) &= \sum_{k=1}^\infty \dfrac{3k+1}{k(k+1)}y^{k+1}\\ -2\ln(1-y)-y\ln(1-y)-2y &= \sum_{k=1}^\infty \dfrac{3k+1}{k(k+1)}y^{k+1} \hspace{3cm}\text{0<y<1}\\ \end{align} now set $y=1-x$ and drop two first terms. So $$\sum_{k=3}^\infty (-1)^k \frac{3k+1}{k(k+1)}(x-1)^k=\color{blue}{-\frac{31}{6}+\frac{13}{3}x-\frac{7}{6}x^2-\frac{2}{1-x}\ln x-\ln x}$$