I want to calculate limn→∞n∏k=1(1+tankn2)
Taking logarithms, it's enough to find
limn→∞n∑k=1log(1+tankn2).
Since limn→∞tanxn2=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
limx→0tanxx=1
and, as shown here
limn→∞1nn∑k=1f(kn)=1∫0f(x)dx
n∑k=1log(1+tankn2)=n∑k=1tankn2⋅log(1+tankn2)1tankn2=n∑k=1kn2⋅tankn2kn2⋅log(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)
(1−ε)(n∑k=1kn2)<n∑k=1log(1+tankn2)<(1+ε)(n∑k=1kn2)
leading to
limn→∞n∑k=1log(1+tankn2)=limn→∞n∑k=1kn2
But then
limn→∞n∑k=1kn2=limn→∞1n(n∑k=1kn)=1∫0xdx=12
and the result follows.
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