Given that ∑ 𝑢𝑘 ∞ 𝑘=1 converges with 𝑢𝑘 > 0, prove that ∑ √𝑢𝑘.𝑢𝑘+1 ∞ 𝑘=1 also converges. Show that the converse is also true if 𝑢𝑘 is monotonic.
∑unun+1\sum \sqrt{u_nu_{n+1}}∑unun+1
since uk is monotonic, then:
un/un+1<1u_n/u_{n+1}<1un/un+1<1
unun+1<un+12u_nu_{n+1}<u_{n+1}^2unun+1<un+12
unun+1<un+1\sqrt{u_nu_{n+1}}<u_{n+1}unun+1<un+1
so, since ∑un+1=∑un+un+1\sum u_{n+1}=\sum u_n+u_{n+1}∑un+1=∑un+un+1 converges, series ∑unun+1\sum \sqrt{u_nu_{n+1}}∑unun+1 converges as well
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