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Does there exist $f(n)>n\ln n$ such that $$\sum_{n=k}^\infty\frac{1}{f(n)}$$ diverges?

If so, is there a maximum such $f(n)$?

Of course the answer to the last question I suppose would eliminate the need for all convergence tests (given that the series is not alternating)!

John Molokach
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    You may be interested in this. – David Mitra Jul 28 '16 at 13:54
  • Liad, no. Series will converge in that case. – John Molokach Jul 28 '16 at 14:00
  • More simply: suppose $\sum a_n$ is infinite with $a_n>0$. Choose $n_1$ so that $a_1+\cdots+a_{n_1}>1$. Replace each of these $a_i$ with $a_i/2$. Now choose $n_2$ so that $a_{n_1+1}+\cdots+a_{n_2}>1$. replace each of these $a_i$ with $a_i/3$. Continue... You'll wind up with a "smaller" divergent series. – David Mitra Jul 28 '16 at 14:04
  • But the infinite sum of $\frac {1}{n\ln n}$ converges so as it's tail. If there exsist such $f$ then by comparison test $\frac {1}{n\ln n}$ must diverge. Am I wrong? – Snufsan Jul 28 '16 at 14:04
  • Snufsan, that sum does not converge. Use the integral test. – John Molokach Jul 28 '16 at 14:05
  • David Mitra, there is no $sup$ of this set? I get the math, but this seems like a paradox. BTW I would still like to have a few examples. Perhaps with $ln(n)$. – John Molokach Jul 28 '16 at 14:07

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By Cauchy's condensation test, all the following series are divergent: $$ \sum_{n\geq n_0}\frac{1}{n\log(n)},\qquad \sum_{n\geq n_0}\frac{1}{n\log(n)\log\log(n)},\qquad \sum_{n\geq n_0}\frac{1}{n\log(n)\log\log(n)\log\log\log(n)} $$ and all the following series are convergent: $$ \sum_{n\geq n_0}\frac{1}{n\log(n)^2},\qquad\! \sum_{n\geq n_0}\frac{1}{n\log(n)\log\log(n)^2},\qquad \!\sum_{n\geq n_0}\frac{1}{n\log(n)\log\log(n)\log\log\log(n)^2} $$ so there is clearly no maximum.
In each case, $n_0$ is chosen in such a way that every term of the corresponding sum makes sense.

Jack D'Aurizio
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  • Sorry, the "clearly" part eludes me. Continue your progression of divergent series ad infinitum. There is no $sup$ of the denominator? it is easy to get a convergent series from any of the series in this set by repeating any factor, i.e. multiplicity 2 or more. But if we don't do that, then....? – John Molokach Jul 28 '16 at 14:28
  • NVM. I see it. Any product of divergent expressions will also diverge - so clearly so $sup$. Thanks. – John Molokach Jul 28 '16 at 14:29
  • I was going to say that you may simply add extra $\log\log\log\ldots$ factors in the denominators of a series in the first line, preserving divergence. – Jack D'Aurizio Jul 28 '16 at 14:31