Is there a function $f : \mathbb{N} \to \mathbb{R}^+$ in $o(1/n)$ s.t. $\Sigma_{i=0}^\infty f(n)$ diverges?
Is there a diverging series in $o(1/n)$
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$\begingroup$
real-analysis
sequences-and-series
asymptotics
divergent-series
2 Answers
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Yes. let $g(n)=\ln\ln n$. Then $g(n+1)-g(n)\approx g'(n)=\frac 1n\cdot\frac1{\ln n}\in o(\frac1n)$. Thus letting $f(n)=g(n+1)-g(n)$, you have your example
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Try $\sum_{i=2}^\infty \frac{1}{n \log(n) }$
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4And use the integral test. – 2012-09-19