OpenStudy (maya):

∑1∞1/(2n+1)(2n−1)

6 years ago
OpenStudy (anonymous):

the sum diverges

6 years ago
OpenStudy (maya):

how did you figure that out?

6 years ago
OpenStudy (anonymous):

wolfram

6 years ago
OpenStudy (maya):

what is the partial sum technique?

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (maya):

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

6 years ago
OpenStudy (maya):

It is not what you have there

6 years ago
OpenStudy (anonymous):

mine is simplified

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

that might help

6 years ago
OpenStudy (maya):

alright

6 years ago
OpenStudy (maya):

the ans i got from my prof= 1/2

6 years ago
OpenStudy (maya):

that means it does converge

6 years ago
OpenStudy (anonymous):

im not sure then sorry

6 years ago
OpenStudy (maya):

np thanks anyways

6 years ago
OpenStudy (anonymous):

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 :)

6 years ago
OpenStudy (maya):

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

6 years ago
OpenStudy (anonymous):

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.

6 years ago
OpenStudy (maya):

got it! thanks :)

6 years ago
OpenStudy (anonymous):

:)

6 years ago
OpenStudy (maya):

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

6 years ago
OpenStudy (maya):

does this converge or diverge?

6 years ago