You are given that $v=0.9$. You are also given a non-level perpetuity that pays the amount of $\frac{1}{k(k+1)}$ at times $k=1,2,3,\ldots$. Find the present value of this perpetuity.
I am given that integrating both sides of the function $1+x+x^2+\cdots=\frac{1}{(1-x)}$ as $|x|\lt 1$ should help solve this. (which is $x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots = -\ln(1-x)+C$)
I get $\frac{v}{1(1+1)} + \frac{v^2}{2(2+1)} + \frac{v^3}{3(3+1)} + \cdots$
I have tried factoring $\frac{v}{2}$ which gives me $1+\frac{v}{3} + \frac{v^2}{6}+\cdots$ to get it into a form that makes the hint useful. Am I missing something?
The present value is $$ PV=\sum_{k=1}^\infty \frac{v^k}{k(k+1)}=\sum_{k=1}^\infty \left(\frac{1}{k}-\frac{1}{k+1}\right)v^k=\sum_{k=1}^\infty \frac{v^k}{k}-\sum_{k=1}^\infty \frac{v^k}{k+1} $$ We have $$ \sum_{k=1}^\infty \frac{v^k}{k}=-\log(1-v) $$ and $$ \sum_{k=1}^\infty \frac{v^k}{k+1}=\frac{1}{v}\sum_{k=1}^\infty \frac{v^{k+1}}{k+1}=\frac{1}{v}\sum_{n=2}^\infty \frac{v^{n}}{n}=\frac{1}{v}\left(\sum_{n=1}^\infty \frac{v^{n}}{n}-v\right)=\frac{1}{v}\left(-\log(1-v)-v\right) $$ So we find $$ PV=\sum_{k=1}^\infty \frac{v^k}{k(k+1)}=\left(\frac{1}{v}-1\right)\log(1-v)+1=0.744157 $$