Is there another representation for the summation: $\sum_{j=1}^{N}\frac{a_j}{(c+a_j)(c+a_j+1)} $, how to reformulate that to keep $c$ out of the sum [closed]

Question

Is there a closed form (without summation) for the summation or at least can I reformulate that so I keep $c$ out of the summation, for example, $c \sum_{n=1}^{N} f(a_n,b_n)$.

$$ \sum_{n=1}^{N}\frac{a_n}{(c+b_n)(c+b_n+1)} $$ where $a_{j},b_{j},c>0$. I have seen some examples in the book of series and integrals:

\begin{gather*} \sum_{k = 1}^n \frac1{[p + (k - 1)q](p + k q)} = \frac n{p(p + n q)} \\ \sum_{k = 1}^n \frac1{[p + (k - 1)q](p + k q)[p + (k + 1)q]} = \frac{n(2p + n q + q)}{2p(p + q)(p + n q)[p + (n + 1)q]}. \end{gather*}

I faced this summation while solving an equation, the final equation was:

$$ \frac{-1}{c^2}+ \sum_{n=1}^{N}\frac{a_n}{(c+b_n)(c+b_n+1)}=0. $$

I was thinking if I could take $c$ out of the summation in any way, so to solve the equation for $c$.

Answer 1

A few words on existence and approximation. Write your equation as
$$ \sum_{n=1}^N \frac{c^2a_n}{(c+b_n)(c+b_n+1)}=1.$$ There is a nice partial fractions decomposition, $$\frac{c^2 }{(c+b )(c+b +1)}=1+\frac{b^2}{c+b}-\frac{(b+1)^2}{c+b+1},$$ which is increasing for positive $c$, as one can check from the derivative. Therefore, every term of the LHS of the above equation is continuous and strictly increasing wrto $c$, with values at $0$ and limit at $+\infty$ $0$, resp. $\sum_{n=1}^Na_n$.

We conclude that there exists exactly one positive solution $c_*>0$ provided $\sum_{n=1}^Na_n>1,$ and no solutions if $\sum_{n=1}^Na_n\le 1.$

Assuming $\sum_{n=1}^Na_n>1,$ you may approximate the solution by monotone iteration. Write e.g. the equation as $f(c)=c$ with $$f(c):= \sum_{n=1}^N \frac{c^{\mathbf 3} a_n}{(c+b_n)(c+b_n+1)}, $$ then since by the above observations $f$ is strictly increasing and $f(c)<c$ iff $c<c_*$, you get a monotone sequence $c_{k+1}:=f(c_k)$ converging to $c_*$ (increasing or decreasing, provided that $c_0<c_*$ or $c_0>c_*$)