Question

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

Answer 1

@ganeshie8

Answer 2

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

Answer 3

what does it converge to

Answer 4

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

Answer 5

1/2 ?

Answer 6

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

Answer 7

That is what @seascorpion1 's simplification converge to

Answer 8

3/2 ?

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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