Calculate $\frac{a_{1}}{3}+\frac{a_{2}}{5}+\frac{a_{3}}{7}+\cdots+\frac{a_{2006}}{4013}$ [closed]

Solution 1:

(I am working with hand and paper, so this may contain mistakes in the calculations. However, the method works)

Consider the rational function $$ F(X) = \sum_{i=1}^{2006} \frac{a_i}{X+i}$$

By definition of the $a_i$'s, one has, for every $j \in [1,2006]$,

$$F(j)= \frac{4}{2j+1}$$

Hence, the polynomial $\displaystyle{P(X) = \left(\prod_{i=1}^{2006}(X+i) \right) F(X)}$ is a polynomial of degree $2005$ which satisfies, for every $j \in [1, 2006]$, $$P(j) = \frac{(j+2006)!}{j!} \times \frac{4}{2j+1}$$

By Lagrange's interpolation, one has $$P(X) = \sum_{j=1}^{2006} \left(\frac{(j+2006)!}{j!} \times \frac{4}{2j+1} \right)\prod_{i = 1\\ i \neq j}^{2006} \frac{X-i}{j-i}$$

In particular, one has

\begin{align*} P \left( \frac{1}{2}\right) &= \sum_{j=1}^{2006} \frac{(j+2006)!}{j!} \times \frac{4}{2j+1} \prod_{i = 1\\ i \neq j}^{2006} \frac{\frac{1}{2}-i}{j-i}\\ &= \sum_{j=1}^{2006} \frac{(j+2006)!}{j!} \times \frac{4}{(2j+1)2^{2005}} \prod_{i = 1\\ i \neq j}^{2006} \frac{1-2i}{j-i}\\ &= \sum_{j=1}^{2006} \frac{(j+2006)!}{j!} \times \frac{4}{2j+1} \frac{4012 !(-1)^{2007-j}}{2^{4011} 2006 !(j-1)!(2006-j)!(2j-1)}\\ \end{align*}

So \begin{align*} F \left( \frac{1}{2}\right) &= \frac{P \left( \frac{1}{2}\right)}{\prod_{i=1}^{2006}(\frac{1}{2}+i)}\\ &= \frac{2^{4012}2006 ! P \left( \frac{1}{2}\right)}{4013 !}\\ &= \sum_{j=1}^{2006} \frac{(j+2006)!}{j!} \times \frac{8}{4j^2-1} \frac{(-1)^{2007-j}}{4013 (j-1)!(2006-j)!}\\ &= \sum_{j=1}^{2006} {2006 \choose j}{j+2006 \choose j} \times \frac{(-1)^{2007-j}8}{4j^2-1} \frac{j}{4013}\\ \end{align*}

So finally, noticing that

\begin{align*} \sum_{i=1}^{2006} \frac{a_i}{2i+1} = \frac{1}{2} F \left(\frac{1}{2} \right) \end{align*}

you get $$\sum_{i=1}^{2006} \frac{a_i}{2i+1} = \frac{4}{4013}\sum_{j=1}^{2006} {2006 \choose j}{j+2006 \choose j} \times \frac{(-1)^{2007-j}j}{4j^2-1}$$

which seems to simplify into

$$\boxed{\sum_{i=1}^{2006} \frac{a_i}{2i+1} = 1-\frac{1}{4013^2}}$$