Monday, 22 December 2014

calculus - limlimitsntoinftyprodlimitsnk=1left(1+tanfrackn2right)



I want to calculate limnnk=1(1+tankn2)



Taking logarithms, it's enough to find
limnnk=1log(1+tankn2).



Since limntanxn2=0, we can combine the Taylor series near 0 of log(1+x) with the taylor series of tanx near 0 to obtain the limit e12.




My question is: is there any nicer way of evaluating this limit?


Answer



Probably not nicer, but still a different way is to use the facts that
limx0tanxx=1


and, as shown here
limn1nnk=1f(kn)=10f(x)dx






nk=1log(1+tankn2)=nk=1tankn2log(1+tankn2)1tankn2=nk=1kn2tankn2kn2log(1+tankn2)1tankn2


Because the part in red 1 when n for any k=1..n, using the definition of the limit, ε,N(ε) s.t. n>N(ε)
(1ε)(nk=1kn2)<nk=1log(1+tankn2)<(1+ε)(nk=1kn2)

leading to
limnnk=1log(1+tankn2)=limnnk=1kn2

But then
limnnk=1kn2=limn1n(nk=1kn)=10xdx=12


and the result follows.


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