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If $a_n$ is a sequence of non-increasing positive numbers, then suppose we already know that $$\sum_p a_p$$ converges, when $p$ runs over the primes, what should be used to prove that $$\sum_n \frac{a_n}{\log{n}}$$ also converges, where $n$ runs over the positive naturals?

And also, how to show the converse is also true?

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Do you mean that the first series is summed over all primes and the second series is summed over all positive integers? –  Bill Cook Oct 16 '11 at 1:31
    
Sounds intuitively plausible, given the Prime Number Theorem. –  Henning Makholm Oct 16 '11 at 1:52
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@robjohn: I call that strictly decreasing - see en.wikipedia.org/wiki/Monotonic_function –  AD. Oct 16 '11 at 10:50
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@robjohn: You just said that monotone decreasing adopts a strict inequality. –  AD. Oct 16 '11 at 12:44
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@AD: there are various usages. Just saying "decreasing" is usually ambiguous. Monotone decreasing is less ambiguous and usually means strictly decreasing, which is $a_{n+1}<a_n$. non-increasing definitely means $a_{n+1}\le a_n$. I have also seen monotone non-increasing also. –  robjohn Oct 16 '11 at 13:22

3 Answers 3

I would use (1) $p_n \approx n\ln n$ and $p_{n+1} < (1+\epsilon)p_n$ for any $\epsilon$ for large enough $n$. This will allow you to handle the step from $a_{p_n}$ to $a_{p_{n+1}}$.

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$$ \sum_n \frac{a_n}{\log n} \leq \sum_n \frac{a_{p_n}}{\log p_n}(p_{n+1}-p_n)$$

Now if you prove that the sequence $ \frac{p_{n+1}-p_n}{\log p_n}$ is bounded, you are done. I guess that this would be easy using the fact that $\frac{p_n}{n \log n} \to 1$ as $n \to \infty$.

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I would be very interested in the easy proof of the boundedness of $\frac{p_{n+1}-p_n}{\log p_n}$. –  robjohn Oct 16 '11 at 14:27
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The gaps between primes are not O(log n). –  zyx Oct 17 '11 at 4:09
    
Ok then. Sorry about that. I will delete the answer. –  Beni Bogosel Oct 17 '11 at 5:57

Going back to the ratio test (http://en.wikipedia.org/wiki/Ratio_test), we have: $$\lim_{n\rightarrow \infty} \sup \frac{a_{p_{n+1}}}{a_{p_{n}}}=R$$ where $R \leq 1$ (it exists because $0 < \frac{a_{p_{n+1}}}{a_{p_{n}}} \leq 1 $, i.e. bounded).

If $R < 1$, then (1) $$\sum_{n}\frac{a_{n}}{\log n}=\sum_{n}\frac{a_{n}\cdot \pi(n)}{n}\cdot \frac{n}{\pi(n)\cdot \log n}< \left ( 1 + \varepsilon \right )\sum_{n}\frac{a_{n}\cdot \pi(n)}{n}$$ Now, assuming: $$p_{n}=n_{0}<n_{1}<...n_{k}<p_{n+1}$$ we have $$a_{p_{n}}\geq a_{n_{j}}, 0\leq j\leq k$$ $$n=\pi(p_{n})=\pi(n_{0})=\pi(n_{1})=...=\pi(n_{k})$$ and $$\frac{a_{p_{n}}\cdot \pi(p_{n})}{p_{n}} \geq \frac{a_{n_{j}}\cdot \pi(n_{j})}{n_{j}}$$ or (2) $$\left ( k+1 \right )\cdot \frac{a_{p_{n}}\cdot \pi(p_{n})}{p_{n}}\geq \sum_{j=0}^{k}\frac{a_{n_{j}}\cdot \pi(n_{j})}{n_{j}}$$ Considering that $k+1=p_{n+1}-p_{n}$ and $p_{n+1}-p_{n} < p_{n}$ (http://en.wikipedia.org/wiki/Bertrand's_postulate) $$a_{p_{n}}\cdot \pi(p_{n}) \geq \sum_{j=0}^{k}\frac{a_{n_{j}}\cdot \pi(n_{j})}{n_{j}}$$ Or ($\pi(p_{n})=n$): $$\sum_{n}\frac{a_{n}}{\log n} < \left ( 1 + \varepsilon \right ) \sum_{n}\frac{a_{n}\cdot \pi(n)}{n}\leq \left ( 1 + \varepsilon \right )\sum_{n}n\cdot a_{p_{n}}$$ Now $$\lim_{n\rightarrow \infty } \sup \frac{(n+1)\cdot a_{p_{n+1}}}{n\cdot a_{p_{n}}}=R< 1$$ It is also worth noting that for $\forall p_{k} \leq n$, $a_{p_{k}} \geq a_{n}$ and as a result: $$\sum_{k=1}^{\pi (n)}a_{p_{k}}\geq \pi (n)\cdot a_{n}$$ Or: $$\lim_{n\rightarrow \infty } \pi (n)\cdot a_{n} \leq \sum_{n}a_{p_{n}}< \infty $$ But if we assume $$\lim_{n\rightarrow \infty } \pi (n)\cdot a_{n}=\gamma >0$$ then from some $n$ we have $\pi (n)\cdot a_{n} > \frac{\gamma }{2}$ or $a_{n} > \frac{\gamma }{\pi (n)\cdot 2}$ or $a_{p_{n}} > \frac{\gamma }{n\cdot 2}$, this will contradict the convergence of the $\sum a_{p}$ and $\gamma =0$. So for $R < 1$ the statement seems to be true.

However for $R=1$, the original condition may not be sufficient. Let's assume $$\lim_{n\rightarrow \infty} \frac{a_{p_{n+1}}}{a_{p_{n}}}=1$$ strictly. Then, (1) becomes: $$\left ( 1 - \varepsilon \right )\sum_{n}\frac{a_{n}\cdot \pi(n)}{n} < \sum_{n}\frac{a_{n}}{log(n)}< \left ( 1 + \varepsilon \right )\sum_{n}\frac{a_{n}\cdot \pi(n)}{n}$$

And (2) becomes: $$\left ( p_{n+1} - p_{n} \right )\cdot \frac{a_{p_{n+1}}\cdot \pi(p_{n})}{p_{n+1}} \leq \sum_{j=0}^{k}\frac{a_{n_{j}}\cdot \pi(n_{j})}{n_{j}} \leq \left ( p_{n+1} - p_{n} \right )\cdot \frac{a_{p_{n}}\cdot \pi(p_{n})}{p_{n}}$$

And: $$\left ( p_{n+1} - p_{n} \right )\cdot \frac{a_{p_{n+1}}\cdot \pi(p_{n})}{p_{n+1}}=\left ( p_{n+1} - p_{n} \right )\cdot \frac{a_{p_{n}}\cdot \pi(p_{n})}{p_{n}} \cdot \frac{p_{n}}{p_{n+1}} \cdot \frac{a_{p_{n+1}}}{a_{p_{n}}}$$

This indicates (considering the assumption) that: $$\sum_{n}\frac{a_{n}}{\log n} < \infty \Leftrightarrow \sum_{n} \left ( p_{n+1} - p_{n} \right )\cdot \frac{a_{p_{n}}\cdot \pi(p_{n})}{p_{n}} < \infty \Leftrightarrow \sum_{n} \frac{p_{n+1} - p_{n}}{\log p_{n}} \cdot a_{p_{n}}< \infty$$

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Unfortunately, the case $r=1$ is where most of the interesting behaviour lies. Also, you seem to be assuming a converse of the ratio test: the limit $r$ might not even exist (although the lim sup is certainly $\le 1$ in this case). –  Erick Wong Jan 28 '13 at 0:56
    
I see, then switching to root test (en.wikipedia.org/wiki/Root_test) should work for the case when $r<1$, considering that $\lim_{n \to \infty }\sqrt[n]{n}=1$ –  rtybase Jan 28 '13 at 9:46

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