Answer to Question #280120 in Real Analysis for abc

Question

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.

Answer 1

$\sum \sqrt{u_nu_{n+1}}$

since u_k is monotonic, then:

$u_n/u_{n+1}<1$

$u_nu_{n+1}<u_{n+1}^2$

$\sqrt{u_nu_{n+1}}<u_{n+1}$

so, since $\sum u_{n+1}=\sum u_n+u_{n+1}$ converges, series $\sum \sqrt{u_nu_{n+1}}$ converges as well