$u_{n}>u_{n+1}>0$ and $u_{2}+u_{4}+u_{8}+u_{16}.....$diverges. prove that $\sum \frac{u_{n}}{n}$ diverges. The only thing i found is that $\left | \frac{u_{2^{n+1}}}{2u_{2^n}} \right |>1$.
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Here's a hint: $$ \underbrace{\frac{u_1}{1} + \frac{u_2}{2}}_{>\,2\cdot u_2/2} + \underbrace{\frac{u_3}{3} + \frac{u_4}{4}}_{>\,2\cdot u_4/4} + \underbrace{\frac{u_5}{5} + \frac{u_6}{6} + \frac{u_7}{7} + \frac{u_8}{8}}_{>\,4\cdot u_8/8} + \cdots. $$ Do you see what I'm doing? You might find information about the Cauchy condensation test useful.
user134824
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2looks similar to the proof of why $\sum \frac{1}{n}$ diverges. – V150 Jun 04 '14 at 02:49
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Exactly. Do you think it converges or diverges? – user134824 Jun 04 '14 at 02:50
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Since it's actually larger than $\frac{u_{2}}{2}+\frac{u_{4}}{2}+\frac{u_{8}}{2}+\frac{u_{16}}{2}+\frac{u_{32}}{2}.....$ which diverges, I think it diverges. – V150 Jun 04 '14 at 03:00
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That's right! (Except you want $u_2$ instead of $u_2/2$.) – user134824 Jun 04 '14 at 03:01
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There is a general theorem if $a_n$ is decreasing both $\sum a_n$ and $\sum 2^n a_{2^n}$ converge or both diverge.
Rene Schipperus
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