Find the coefficient of $x^{29}$ in the given polynomial. The polynomial is :
$$
\left(x-\frac{1}{1\cdot3}\right) \left(x-\frac{2}{1\cdot3\cdot5}\right) \left(x-\frac{3}{1\cdot3\cdot5\cdot7}\right) \cdots \left(x-\frac{30}{1\cdot3\cdot5\cdots61}\right)
$$
What I've done so far : The given polynomial is an expression of degree $30$. Hence, the coefficient of $x^{29}$ will be the negative of the sum of the roots. But the resulting sum is too complicated to handle and I think I'm doing it wrong. 
 A: HINT: The sum $S_k $ given by 
$$\sum_{n=1}^{k} \dfrac {n}{(2n+1)!!} $$
for  $k=1,2,3....$ gives 
$$S_1=\frac {1}{3} \\ =\frac {1}{2}-\frac {1/2}{ 1 \cdot 3}$$
$$S_2=\frac {1}{3} + \frac {2}{15} =\frac {7}{15} \\= \frac {1}{2}-\frac {1/2}{1 \cdot 3 \cdot 5 }$$
$$S_3=\frac {1}{3} + \frac {2}{15} +\frac {3}{105} =\frac {52}{105} \\= \frac {1}{2}-\frac {1/2}{1 \cdot 3 \cdot 5 \cdot 7 }$$
You can use induction to confirm this pattern. 
A: The coefficient is $-\sum_{i=1}^{30}\frac {i}{\prod_{j=1}^i (2j+1)} $
Claim: let $M_n = \prod_{j=1}^n (2j+1) $ then $\sum_{i=1}^{n}\frac {i}{\prod_{j=1}^i (2j+1)}=\frac {M_n-1}{2M_n} $.
Proof by induction:
Base step: $1/1*3 = \frac {3-1}{2*3} $.
Induction step: Suppose $\sum_{i=1}^{n}\frac {i}{\prod_{j=1}^i (2j+1)}=\frac {M_n-1}{2M_n} = \frac {M_n-1}{2M_n}$.  Then for $n+1$:
$\frac {M_n-1}{2M_n}+\frac {n+1}{M_n*(2(n+1)+1)}=\frac {(M_n-1)(2 (n+1)+1)}{2M_n*(2 (n+1)+1)}+\frac { (n+1)*2}{2M_{n+1}} $
$= \frac {(M_n-1)(2 (n+1)+1)+ 2(n+1)}{2M_{n+1}}=$
$\frac {M_{n+1}-2 (n+1)-1+2(n+1)}{M_{n+1}}=$
$\frac {M_{n+1}-1}{M_{n+1}}=$
So the coefficient is:
$-\frac {1*3*5*7*....*61-1}{2(1*3*5*7*....*61)} $
A: Let
$p_m(x)
=\prod_{k=1}^{m} (x-\dfrac{k}{\prod_{j=1}^{k+1}(2j-1)})
$
Since,
in the usual way,
$\begin{array}\\
\prod_{j=1}^{k+1}(2j-1)
&=\dfrac{\prod_{j=1}^{2k+1}j}{\prod_{j=1}^k(2j)}\\
&=\dfrac{(2k+1)!}{2^kk!}\\
\end{array}
$
$p_m(x)
=\prod_{k=1}^{m} (x-\dfrac{k2^kk!}{(2k+1)!})
$
The coefficient of $x^{m-1}$
is $(-1)^{m-1}$ times the sum of the roots,
which is
$\begin{array}\\
\sum_{k=1}^m \dfrac{k}{\dfrac{(2k+1)!}{2^kk!}}
&=\sum_{k=1}^m \dfrac{k2^kk!}{(2k+1)!}
\\
\end{array}
$
According to Wolfy,
this sum approaches
$\frac12$ as $m \to \infty$.
Calling this sum
$s(m)$,
then
$s(2:5)
=(7/15, 52/105, 472/945, 5197/10395)
$.
To see how close
this is to $\frac12$,
$\frac12-s(2:5)
=(1/35, 1/210, 1/1890, 1/20790)
$.
For $m=30$,
$\frac12-s(30)
=\dfrac1{3564303977319726652772203331132409634218750}
$.
Note that
$s(m)
=\sum_{k=1}^m \dfrac{k2^kk!}{(2k+1)!}
=\dfrac1{(2m+1)!}\sum_{k=1}^m k2^kk!\dfrac{(2m+1)!}{(2k+1)!}
$
which explains,
to a certain extent,
that denominator.
Im particular,
all its prime factors do not exceed
$2m+1$.
A: HINT:$$\underbrace{1.3,1.3.5,...,1.3.5....61}_{30-terms} $$so 
$$\underbrace{(x-\frac{1}{1.3})(x-\frac{2}{1.3.5})(x-\frac{3}{1.3.5.7})......(x-\frac{30}{1.3.5.....61})}_{30-terms }$$we have 
,to obtain $x^{29}$ 
we have
when you want 
$x^{29}$ from $(x+a_1)(x+a_2)(x+a_3)...(x+a_{30})$ you have to find this
$$(x+0)(x+0)(x+0)...(x+0)(a_{30})+\\(x+0)(x+0)(x+0)...(a_{29})(x+0)+\\(x)(x)(x)...(a_{28})(x+0)(x+0)+\\
(x+0)(x+0)(x+0))...(a_{27})(x+0)(x+0)(x+0)+\\\vdots\\
(x+0)(a_{2})...(x+0)(x+0)(x+0)+\\
(x+a_1)(a+0)...(x+0)(x+0)(x+0)$$
