Sum of n terms of this series $\frac{1}{1.3} + \frac{2}{1.3.5} +\frac  {3}{1.3.5.7} + \frac{4}{1.3.5.7.9}........ n  $ Terms.
I Know the answer to this problem but I couldn't find any proper way to actually solve this question. I thought  the denominators were the product of n odd natural numbers. So I wrote the nth term as:
$\frac{2^{n+1}(n+1)!.n }{[2(n+1)]!}$
But I don't know if that's correct or even what to do next.?

The $n$th term has $n$ as its denominator, and the odd factors in $(2n+1)!$ as its numerator. Or, if we multiply both by the even factors, which are $\prod_{i=1}^n 2i=n!2^n$, the denominator becomes $(2n+1)!$. The $n$th term is therefore $\dfrac{n! n2^n}{(2n+1)!}$. (The formula you've obtained is also correct, by the way; you've just added a factor of $2(n+1)=2n+2$ into the two parts of the fraction.)
Alternatively, we could leave the numerator and denominator as they were originally, and write the $n$th term as $\dfrac{n}{(2n+1)!!}$, where the subfactorial $k!!$ is defined to only include the factors from $1$ to $k$ of the same parity as $k$. (In particular, the rescaling I suggested above used a factor of $(2n)!!$.)
As for the sum of the first $n$ terms, I think you can prove by induction it's $\frac{a_n}{2a_n+1}$ with $2a_n+1=(2n+1)!!$.
