The limit:limn→∞((3nn)(2nn))1n
What I did was put limit = L. Then,
log(L)=limn→∞1n⋅n−1∑r=0log(3−rn2−rn)=∫10log(3−x2−x)dx=log(2716)
Is this aproach correct? Is there other method.
Edit: I have corrected the expression for the limit.
Answer
limn→∞((3nn)(2nn))1n=limn→∞(n!(3n)!(2n)!2)1n=limn→∞exp(ln(n!(3n)!(2n)!2)n)=limn→∞exp(ln((√2πn(ne)n)(√2π(3n)(3ne)3n)(√2π(2n)(2ne)2n)2)n)=limn→∞exp(ln(36n+1224n+1)n)=limn→∞exp(ln(e(6n+12)ln3e(4n+1)ln2)n)=limn→∞exp(ln(2716)−ln(43)2n)=2716
Solved with Stirling approximation
x!≈√2πx(xe)x, for x→∞
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