Simplify formula by removing sum from it

Question

I have a question about the following formula, I just need to write it without a sum with some formula convertion(simplify it). As for now I'm stuck, I don't know if it is possible for now, but hope it is. $$ \sum_{i=1}^l \frac{b_i^2}{1+AB_i}, \quad \text{where }B_i=\sum_{\textstyle j= \, i}^L b_{\textstyle j}, \ \text{and } \ l\leq L $$ I tried to lead to a common denominator, make some assumptions to simplify the formula, but for now I'm stuck, any help is appreciated. $$ \sum_{i=1}^l \frac{b_i^2}{1+AB_i}= \frac{b_1^2}{1+A(b_1+...+b_L)} + \frac{b_2^2}{1+A(b_2+...+b_L)}+...+\frac{b_l^2}{1+A(b_l+...+b_L)} $$

Answer 1

There are $2D$ sums in the expression, which can be transformed, using formulas for the negative binomials

$$\frac1{(1-x)^{m+1}} = \sum_{n=m}^\infty \binom nm x^{n-m}$$ which allow to write according polynomial formulas, wherein the polynomial coefficients equal to the binomial coefficients production.

Simplification of the polynomial formulas requires additional assumptions about given coefficients (positive? sorted?$\dots$ Are $Ab_i < 1?$ etc.)

Partially, if $b_i\in(0,1)$ and $0<A<<1,$ then the approximation $$1+A\sum b_i\approx \prod(1+Ab_i)$$ allows to obtain suitable common denominators.

Another essential advantages has the approximation $$1+A\sum b_i \approx e^{A b_i}.$$

If the sequence $\{b_l\}$ increases and $0<A<<1$ or $L>>l,$ i.e. if $1+AB_l > A(B_1-B_l),$ then looks correct the next approach.

Denote $D_i=1+AB_i,\quad D = D_l,\quad \alpha\,=\dfrac A{D_l},$ then \begin{align} &D_{l-1}= 1+AB_{l}+Ab_{l-1} = (1+AB_l)\left(1+\dfrac{A}{1+AB_l}b_{l-1}\right) \approx D e^{\alpha\, b_{l-1}},\\[4pt] &D_{l-2}\approx D_{l-1} e^{\alpha\,b_{l-2}} = De^{\alpha\,(B_{\,l-2}-B_{\,l})},\dots,\\[4pt] &D_{i}\approx D_{i+1} e^{\alpha\,b_i} = De^{\alpha\,(B_{\,i}-B_{\,l})},\dots,\\[4pt] &D_{1}\approx D_{2} e^{\alpha\,b_1} = De^{\alpha\,(B_{\,1}-B_{\,l})},\\[4pt] &\sum_{i=1}^l\dfrac{b_i^2}{1+AB_i} \approx \sum_{i=1}^l\dfrac{b_i^2}{D e^{\alpha\,(B_{\,i}-B_{\,l})}}, \end{align}

$$S(\alpha) = \sum_{i=1}^l\dfrac{b_i^2}{1+AB_i} \approx \dfrac1{D}\sum_{i=1}^l b_i^2e^{-\alpha S_i},\quad\text{where}\quad S_i=\sum_{j=i}^{l}b_j.$$

Obtained expression looks more suitable for the further detalizations. But I cannot estimate real progress.

On the other hand,

\begin{align} &D_{l-1}= 1+AB_{l}+Ab_{l-1} = (1+AB_l)\left(1+\dfrac{A}{1+AB_l}b_{l-1}\right),\\[4pt] &D_{l-2}= 1+AB_{l}+Ab_{l-1}+Ab_{l-1} = D\left(1+\alpha b_{l-2}+\alpha b_{l-1}\right)\approx D\left(1+\alpha b_{l-2}\right)\left(1+\alpha b_{l-1}\right),\dots,\\[4pt] &D_{i}\approx D\left(1+\alpha b_{i}\right)\left(1+\alpha b_{i+1}\right)\dots\left(1+\alpha b_{l-1}\right),\dots,\\[4pt] &D_{1}\approx D\left(1+\alpha b_{1}\right)\left(1+\alpha b_{2}\right)\dots\left(1+\alpha b_{l-1}\right),\\[4pt] \end{align}

$$S(\alpha) \approx \dfrac1{D}\sum_{i=1}^l b_i^2T_i,\quad\text{where}\quad T_l=\dfrac1D,\quad T_i = \dfrac {T_{i+1}}{1+\alpha b_i}.$$