Evaluate the infinite product limn→∞∏k=2n(1−1k2)

Rudy Koch

Rudy Koch

Answered

2022-01-24

Evaluate the infinite product
limnk=2n(11k2)

Answer & Explanation

becky4208fj

becky4208fj

Expert

2022-01-25Added 10 answers

Note that
11k2=(11k)(1+1k)=k1kk+1k=akak1
with ak=k+1k, hence this is a telescoping product, i.e.
k=2n(11k2)=a2a1a3a2anan1=ana1=n+12n
ocretz56

ocretz56

Expert

2022-01-26Added 16 answers

Let g(k)=k1k. Then this product is:
k=2ng(k)g(k+1)
which is a telescoping product, and thus equal to g(2)g(n+1)
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

(11(k1)2)(11(k)2)(11(k+1)2) =k(k2)(k1)(k+1)k(k+2)(k1)2k2(k+1)2=(k2)(k+2)(k1)(k+1) Observe that (11(k)2) is cancelled out by the previous & the next term except for the extreme terms, the 1st & the last term leaving behind the 1st part of the 1st term =212 and the 2nd part of the last term =n+1n

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?