ans- .

FIRST CALCULATE THE Tnth TERM

IT IS

n/((2n+1)!)

just sum it out to find answer

sigma(n/(2n+1)!)

sigma(.5*2n/(2n+1)!)

i.e

.5*sigma((2n+1-1)/(2n+1)!)

.5*sigma(1/(2n-1)!+1/(2n+1)!)

.5*(1/1!-1/3!+1/3!-1/5!..........-1/(2n+1)!)

cancelling we get

0.5*(1-1/(2n+1)!)

that is the answer

the question is not toughest as you say

we both are of same age .

Hi Satyaram,

That is because, nobody gets the r-th term of the series right in this.

The r-th teerm in the series is

T_r = r(r!)2^r/(2r+1)!.

And this is not quite easy to sum. You try to sum this.

Dont ask for the solution. It is a good summation question to do on your own.

Regards,

Ashwin (IIT Madras).

Dude,

When you look at some particular term, say for example the 4th term which is

4/(1.3.5.7.9)

the obvious thing is the Nr is "r"

now comes the tricky part, the Dr (the number of terms in the Dr varies, and are products of odd numbers)

So you have 1.3.5.7.9 - now to represent this in mathematicl notation you need to have (1.2.3.4.5.6.7.8.9)

So 1.3.5.7.9 = (1.2.3.4.5.6.7.8.9)/(2.4.6.8) = (9!)/2⁴4!

so for the r-th term the Dr is (2r+1)!/2^rr!

And hence you get the Tr.

Regards,

Ashwin (IIT Madras).

But then that is not the toughest part.

The tougher part is to sum it. :-)

Regards,

Ashwin (IIT Madras).

Hey this is what was my answer.

I admit my answer is wrong but what is wrong with my nth term

Harikrishnan,

The n-th term is not

n/(2n+1)!. For eg put n=2, your expression will give 2/5! = 2/(1.2.3.4.5). But in the question the 2nd term is 2/(1.3.5).

Ok Satyaram,

For of all there is no solution to your question anywhere above, for you to give a simpler solution :-)

And more over the solution has not been typed!!!.... it has been copy-pasted :-)

That solution is the only way to solve it.

Nice that you got it.

Regards,

Ashwin (IIT Madras).

Dude,

That is the only solution, as I had stated previously.

Do the same thing with the r-th term [ie the general term of the series, so that it will be over with just one term]

If you note

Tr = r(r!)2^r/(2r+1)!

    = 1/2 [ (r-1)!2^r-1/(2r-1)! - r!2^r/(2r+1)! ]

    = 1/2 [ t_r-1 - t_r ]

So, the summation upto infinite terms = 1/2

And summation upto n-terms is 1/2 [ 1 - n!2ⁿ/(2n+1)! ]

Regards,

Ashwin (IIT Madras).

omg i am sorry by (2n+1)! i mean you have to substitue 1 then 2 and so on to n and not substituting n and calulating the factorial.

ok now it is okkk is nt it

or if confusion persists 

just replace (2n+1)!(at all places where you see it) with ∏(2n+1) where ∏ denotes product just like sigma giving n values from 1 to n

then the final answer would be

.5(1-1/(∏(2n+1)))

or

.5(1-(2^n)*n!/(2n+1)!)

hope you all are clear with it

sorry for any misconceptions 

Harikrishnan,

Exactly. Now the change is fine.

Though your method was right, your final answer was wrong just because you had mis-interpreted the n-th term of the series.

Regards,

Ashwin (IIT Madras).