How to study convergence of string #a_n=(1-1/(2^2))(1-1/(3^2))...*(1-1/(n^2))#?,with #n>=2#.

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Answer 1 · 2017-05-27T20:40:47

See below.

We have #1-1/k^2 = (1-1/k)(1+1/k)=(k-1)/k(k+1)/k#

#a_n=(1/2 3/2)(2/3 4/3) cdots ((k-1)/k (k+1)/k)(k/(k+1)(k+2)/(k+1))cdots((n-1)/n(n+2)/n) = 1/2(n+2)/n# then

#lim_(n->oo) a_n = prod_(k=2)^n (1-1/k^2) = 1/2#

How to study convergence of string #a_n=(1-1/(2^2))*(1-1/(3^2))*...*(1-1/(n^2))#?,with #n>=2#.