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Question Number 209308 by Erico last updated on 06/Jul/24
Donnerl′equivalence´simpledeIn=∫01tntn−t+1dt
Answered by mathzup last updated on 07/Jul/24
t=x1n⇒In=∫01xx−x1n+1×1n×x1n−1dx=1n∫01x1nx+1−x1ndx=1n∫01fn(x)dxwehavelimn→∞∫01fn(x)dx=∫01dx1+x=ln2⇒In∼ln2n(n→+∞)
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