Non-increasing sequence of positive real numbers with prime index 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?
 A: 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 $a_{p_{n+1}}$.
A: $$ \sum_n \frac{a_n}{\log n} \leq \sum_n \frac{a_{p_n}}{\log p_n}(p_{n+1}-p_n)$$
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 $n \to \infty$.
A: 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:
$$\sum_{n}\frac{a_{n}}{\log n} < \infty \Leftrightarrow \sum_{n} \left ( p_{n+1} - p_{n} \right )\cdot \frac{a_{p_{n}}\cdot \pi(p_{n})}{p_{n}} < \infty \Leftrightarrow \sum_{n} \frac{p_{n+1} - p_{n}}{\log p_{n}} \cdot a_{p_{n}}< \infty$$ 
