Compute limn→∞∫π20n∑k=1(sinx)kdx.
I tried writing limn→∞∫π20n∑k=1(sinx)kdx=limn→∞∫π201−(sinx)n+11−sinxdx, but I don't know how to continue form here.
I know that the limit diverges to ∞, but I don't know how to prove it.
Answer
The wanted limit is +∞. Indeed, by letting Ik=∫π/20(sinx)kdx we have that {Ik}k≥1 is a decreasing sequence, but it is also log-convex by the Cauchy-Schwarz inequality, and by integration by parts
I2k≥IkIk+1=π2(k+1)
such that
n∑k=1Ik≥√π2n∑k=11√k+1≥√π2n∑k=12(√k+2−√k+1)=√2π(√n+2−√2).
No comments:
Post a Comment