Question

show that:
(1^2)/2! + (2^2)/3! + (3^2)/4! + (4^2)/5! +...... = e-1

Answer 1

use taylor series
for e^x

Answer 2

you'll get it

Answer 3

put x=1

Answer 4

@vamgadu i have no idea regarding taylor series

Answer 5

taylor series expansion of e^x is \[1+x+x ^{2}/2!+x^{3}/3!+....\]
now put x=1 in it

Answer 6

I don't get the solution that way.. can you try it?

Answer 7

no no i am wrong waitno i am wrong wait

Answer 8

is it e-1 or e^-1??

Answer 9

It is not e^-1.. its e - 1

Answer 10

kk

Answer 11

\[n^{2}/(n+1)!=(1/(n+1)!)-(1/(n)!)+(1/(n-1)!)\]

Answer 12

use it
i think its pretty straight forward now

Answer 13

try this way!!
\[ ((1^2)-1)/2! + ((2^2)-1)/3! + ((3^2)-1)/4! + ((4^2)-1)/5! +...... + 1/2! + 1/3! +1/4! + .. \]

Answer 14

\[ \frac {n^2} {(n+1)!} = \frac 1{(n-1)!} + \frac 1{(n+1)!}-\frac 1{n!} \]

You should be able to do it now.

Answer 15

note that \[\sum_{1}^{\inf} 1/n! =e-1\]

Answer 16

@FoolForMath i get \[n ^{2}/(n+1)! =1/(n-1)! - 1/((n+1)(n-1)!) \] how did you get that result you just posted?

Answer 17

take lcm!!

Answer 18

If you do it right, you will get the right answer ;)

Answer 19

Anyways, if you expand my expression , you will see it's converging to \( e-1\) nicely.

Answer 20

i dont think you need to expand at all
as we are summing each term to infinity,
each term give us an e,e-1. and e-2. substitute these in place of summation of reciprocals of (n-1)!,(n)!,(n+1)! respectively. you'll get the answer

Answer 21

Oh yes that's what I meant. I mean I am too fool to see the e series without expanding few terms :P

Answer 22

\[\sum_{1}^{\inf}1/(n-1)!=e and \sum_{1}^{\inf}1/(n+1)!=e-2\]

Answer 23

Yes man I know.

Answer 24

dont blame your self FFm
i know you are a stud. But even the brightest of people make mistakes

Answer 25

that was meant for @ujjwal

Answer 26

lol, I am no stud :)

Answer 27

i know you are

Answer 28

thanks everyone!