<s>[INST] 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.?
 [/INST] the general term is given by $$a_n=\frac{n}{1*3*5....(2n+1)}$$
$$a_n=\frac{\frac{2n+1-1}{2}}{1*3*5....(2n+1)}$$
$$a_n=\frac{1}{2}*({\frac{2n+1}{1*3*5....(2n+1)}-\frac{1}{1*3*5...(2n+1)})}$$
$$a_n=\frac{1}{2}*({\frac{1}{1*3*5....(2n-1)}-\frac{1}{1*3*5...(2n+1)})}$$
for $n=1$ $$\frac{1}{2}*({\frac{1}{1}-\frac{1}{1*3})}$$
for $n=2$ $$\frac{1}{2}*({\frac{1}{1*3}-\frac{1}{1*3*5})}$$
for $n=3$ $$\frac{1}{2}*({\frac{1}{1*3*5}-\frac{1}{1*3*5*7})}$$
thus $$S_n=\frac{1}{2}*(1-\frac{1}{1*3*5*7*....(2n+1)})   $$
$$\lim_{n\to\infty} S_n=\frac{1}{2}$$
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