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[Edited to fix typo]

Is there a precise formulation for when the sum

$$ \sum_{i=2}^{\infty}\frac{1}{i \cdot f(i)} $$ converges, in terms of the function $f$? Assume that $f$ is smooth and monotonically increasing.

If $f(i) \gtrsim i^c$ for any $c>0$ then we know it converges. If $f(i)$ is a constant then we know it doesn't. We can try functions in between. For example setting $f(i) = 2^{\sqrt{\log(i)}}$ makes the sum converge but setting $f(i) = \log(i)$ makes it diverge according to Wolfram Alpha

There are of course a lot of functions so it might be hard to write a full classification. How about if we only including elementary functions that, for example, use only powers and logs?

Update. Is something like the following conjecture true? Consider $\sum_{i=\ell}^{\infty}\frac{1}{i \cdot f(i)}$ and set $\ell$ to be the smallest positive integer so that $f(\ell) >0$. The sum converges if and only if there exists $c>0$ such that $f(i) \gtrsim c \log(i)\log{\log(i)}\log{\log{\log(i)}}\dots$ where the $\log$ is applied an (as yet) unknown but fixed number of times.

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3  
Am I missing something, or the series is just $\sum_{i=2}^\infty\frac1{f(i)}$? –  Dennis Gulko Jan 9 '13 at 17:41
    
@Majid Generally not good to use $f'$ for "some other function," since it carries the derivative connotation a lot of the time. –  Thomas Andrews Jan 9 '13 at 17:55
    
Aargh! That's just a typo! –  user55085 Jan 9 '13 at 18:01
    
What's the right term? –  user55085 Jan 9 '13 at 18:06
    
With $f(i)=\log(i)$, this series does not converge. The integral test can be used to prove that. –  alex.jordan Jan 9 '13 at 18:10

2 Answers 2

up vote 1 down vote accepted

Since $f$ is monotonically increasing, we can use the integral test.

Define repeated composition by $$ f^{\circ0}(x)=x\quad\text{and}\quad f^{\circ k+1}(x)=f\circ f^{\circ k}(x) $$

Note that if $$ f_n(x)=\prod_{k=1}^n\log^{\circ k}(x) $$ then, for $n>0$, $$ \begin{align} \int_{\exp^{\circ n}(1)}^\infty\frac{\mathrm{d}x}{xf_n(x)\log^{\circ n}(x)^a} &\stackrel{\hphantom{\text{induction}}}=\int_{\exp^{\circ n}(1)}^\infty\frac{\mathrm{d}\log(x)}{f_n(x)\log^{\circ n}(x)^a}\\ &\stackrel{\substack{x\mapsto e^x\\\hphantom{\text{induction}}}}=\int_{\exp^{\circ n-1}(1)}^\infty\frac{\mathrm{d}x}{xf_{n-1}(x)\log^{\circ n-1}(x)^a}\\ &\stackrel{\text{induction}}=\int_1^\infty\frac{\mathrm{d}x}{x^{a+1}}\\ &\stackrel{\hphantom{\text{induction}}}=\frac1a \end{align} $$ Therefore, for all $n\ge0$, $$ \int_{\exp^{\circ n}(1)}^\infty\frac{\mathrm{d}x}{xf_n(x)} $$ diverges, yet for any $a>0$, $$ \int_{\exp^{\circ n}(1)}^\infty\frac{\mathrm{d}x}{xf_n(x)\log^{\circ n}(x)^a} $$ converges. As far as logs and powers go, these border convergence/divergence pretty closely.

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Intriguing. The sum seems such a standard question I feel sure someone must have considered it before. –  user55085 Jan 10 '13 at 8:58

Extrapolating off of Marvis's work in the comments section, we see that if $b(n)$ is similar to $\dfrac{1}{\log(n)^{k}}$ for a fixed $k\ge 2$, ($k$ exponent here) then the series will converge. Furthermore, if $b(n)$ is similar to $\dfrac{1}{\log(n)}$ we observe divergence. Consider extending this process to a fixed $j$ so that if $b(n)$ is similar to $\dfrac{1}{\log n\log\log n \log\log\log n ...\log^{(j)} n}$, we have divergence. These are mere examples. In general,
\begin{align*} \sum_{n=2}^{\infty} \log(n)(b(n) - b(n+1)) = \sum_{n=2}^{\infty} \log(n)b(n) - \sum_{n=2}^{\infty} \log(n)b(n+1) \end{align*} If $b(n)$ is a non-decreasing function then
\begin{align*} b(n) \le b(n+1) \Rightarrow \\ \mbox{ For $n\ge2$ we have } \log(n)b(n) \le \log(n)b(n+1) \Rightarrow \\ \log(n)b(n)-\log(n)b(n+1) \le 0 \\ \end{align*} Now $b'(n) \ge 0$.
If $b'(n) = 0$, then the entire sum is $0$. (Uninteresting)
If $b'(n) > 0$ (Increasing) and $b''(n)>0$ (Second Derivative also increasing), then for $n >>2$
$\log(n)b(n)-\log(n)b(n+1)$ ~ $-\log(n)b(n+1)$
$b(n) -b(n+1)$ can get arbitrarily large because the vertical distance between $b(n)$ and $b(n+1)$ approaches $\infty$. This is because for any interval, $[n,n+1]$, we have this by the MVT. So it would stand within reason that the sum would tend to $-\infty$.
What about when $b''(n) < 0 $ but $b'(n) > 0 $ $\forall n$ ? Well the derivative is decreasing implying that for any interval $[n,n+1]$, the vertical distance between $b(n)$ and $b(n+1)$ is decreasing by MVT, so the sum would eventually converge.

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You'd better use $n$ instead of $i$. ;-) –  B. S. Jan 9 '13 at 18:10
    
@BabakSorouh I'm not sure where? And pardon my asking but are you Iranian? –  Rustyn Jan 9 '13 at 19:08
    
RustynYazdanpour You are right that it can't go on forever. So it is true for some fixed $k$? –  user55085 Jan 9 '13 at 19:34
    
@RustynYazdanpour: Although, we are working on series and there is not difficulty to use $i$ rather than $n$; it would be better use $n$. That is it. Like your answer. +1 Any way yes, I am. ;-) –  B. S. Jan 10 '13 at 8:12
    
@BabakSorouh Me too. I have a relative named Babak. –  Rustyn Jan 10 '13 at 14:03

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