Re: [分析] 初微(59)
※ 引述《plover (>//////<)》之銘言：
: Show that the convergence of Σ a_n, where a_n > 0 for all n
: implies the convergence of Σ{(a_n)^(1/2)}/n.
[(a_n)^(1/2) - 1/n]^2 = a_n + 1/n^2 - 2[(a_n)^(1/2)]/n ≧ 0
a_n + 1/n^2 ≧ 2[(a_n)^(1/2)]/n
By assumption, the fact Σ1/n^2 conv. and Comparison test
we know Σ{(a_n)^(1/2)}/n conv.
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