I tried using Stirling's approximation and d'Alambert's ratio test but can't get the limit. Could someone show how to evaluate this limit?
Answer
Use equivalents: n√n!n∼∞(√2πn)1nn⋅ne=1e(2πn)12n Now ln(2πn)12n=lnπ+ln2n2n→n→∞0, hence n√n!n∼∞1e.
I have injection f:A→B and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...
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