In good conditions, when $\{u_n\}$ is monotonically increasing, it's not hard to prove that $$\sum_{i=1}^n u_i \ge [\frac{n+1}{2}]u_[\frac{n+1}{2}]$$Consequently we obtain that $$\frac{n}{\sum_{i=1}^n u_i} \le \frac{n}{[\frac{n+1}{2}]u_[\frac{n+1}{2}]} \le \frac{2}{u_[\frac{n+1}{2}]}$$ Obviously series $\sum_{n=1}^\infty \frac{2}{u_[\frac{n+1}{2}]}$ is convergent, hence series $\sum_{n=1}^\infty \frac{n}{\sum_{i=1}^n u_i}$ is convergent. However, when $\{u_n\}$ is not so good, I think probably adjusting the order of $\{u_n\}$ is a way. I was stuck here. Could someone help me?
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Do you mean to say in the first line that $$\sum_{i=1}^nu_i\geq \left(\frac{n+1}{2}\right)u_{\frac{n+1}{2}}?$$ Btw, what’s your goal? What are you trying to achieve? – insipidintegrator Sep 30 '22 at 09:33
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2Also: The series $\sum\limits_{n=1}^\infty \frac n{\frac1{a_1}+\frac1{a_2}+\dotsb+\frac1{a_n}}$ is convergent – Martin R Sep 30 '22 at 11:05