Question

∑1∞1/(2n+1)(2n−1)

Answer 1

the sum diverges

Answer 2

how did you figure that out?

Answer 3

wolfram

Answer 4

what is the partial sum technique?

Answer 5

if you look at it though, \[\sum_{0}^{infinity} \left( 2n-1 \right)/ (2n+1) \] is always increasing so it doesn't converge

Answer 6

its 1/ (2n+1)*(2n-1)

Answer 7

It is not what you have there

Answer 8

mine is simplified

Answer 9

http://www.wolframalpha.com/input/?i=sum+from+1+to+infinity+1%2F%282n%2B1%29%282n%E2%88%921%29

Answer 10

that might help

Answer 11

alright

Answer 12

the ans i got from my prof= 1/2

Answer 13

that means it does converge

Answer 14

im not sure then sorry

Answer 15

np thanks anyways

Answer 16

the answer is half:
\[.5*((2n+1)-(2n-1))\div((2n+1)(2n-1))\]
se i've written the numerator which results in 2 and hence i divided by 2.
now wen u run n from 1 to infinity, only 1 remains and the rest gets cancelled telescopically , hence the answer half :)

Answer 17

can you explain this please? how do i use partial sums for telescoping series?

Answer 18

ok,multiply and divide numerator and denominator by 2. now write this numerator 2 as (2n+1)-(2n-1).
doing this, u split the sum as partial fractions , and substitue n=1,2,3,... u'll see that only 1 from the first term remains whilst rest all gets cancelled. this cancellation is called telescopic cancellation.

Answer 19

got it! thanks :)

Answer 20

:)

Answer 21

\[\sum_{1}^{\infty} n ^{2} e ^{-2n}\]

Answer 22

does this converge or diverge?