$$1 + \frac{1}{4} + \frac{1\cdot 3}{4\cdot 8 } + \frac{1\cdot 3 \cdot 5}{4\cdot 8 \cdot 12} + \cdots$$

Question

parthkohli · Answer

@ganeshie8

seascorpion1 · Answer

Your series simplifies to:$$\frac{(2n+1)!!}{(4n)!!}$$ for n >=0

parthkohli · Answer

what does it converge to

seascorpion1 · Answer

I'm a bit rusty on this topic but I'll be back in 5mins with an explanation.

dls · Answer

1/2 ?

parthkohli · Answer

nahi re, the first term itself is greater than 1/2.

dls · Answer

That is what @seascorpion1 's simplification converge to

dls · Answer

3/2 ?

parthkohli · Answer

No, the answer is $\sqrt 2$. You should recognise a Maclaurin series in there.

seascorpion1 · Answer

Sorry I couldn't help in time but I learned from your answer.

seascorpion1 · Answer

Are you sure, I get$$\frac{5\sqrt[8]{e}}{4}$$when I used wolfram to check it.

dls · Answer

I guess they are almost the same..accurate upto 2 decimal places

dan815 · Answer

|dw:1449420019624:dw|