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In the book "Function Theory of a Complex Varible" i'm having trouble verifying my proposed proof of the conjecture in $(1.)$

$(1.)$

If $b_{n} > 1$ for all n, then:

$$\prod_{n}b_{n}$$

converges if and only if:

$$\sum_{n}\log(b_{n}) < \infty$$

Proposition $(1.1)$:

Let $a,b \in \mathbb{N}$ with a < b and let $f: \, [a,b] \rightarrow \mathbb{R}$ be a monotonic function on [a,b]. There exists a real number $\theta=\theta(a,b) $ such that $-1 \leq \theta \leq 1$ and such that

$(1.1)$ $$\sum_{a < n \leq b} f(n) = \int_{a}^{b}f(t)dt + \theta(f(b)-f(a)).$$

Lemma:

My initial attack on $(1.)$ begins with making the following observation:

$$\log \prod_{n}b_{n} = \sum_{n}^{}(\log(log(b_{n})) < \infty$$

With further cleanup it is evident that our original conjecture now becomes the following:

$$\log b_{n} = b_{2} \cdot b_{3} \cdot b_{4} \cdot b_{5} \cdot b_{6} \cdot b_{7} \cdot b_{8} \cdot b_{9} \cdot \cdot \cdot \cdot b_{n} = \sum_{n}(log(log(b)-log(log(n) < \infty$$

$$\log b_{n} = b_{2} \cdot b_{3} \cdot b_{4} \cdot b_{5} \cdot b_{6} \cdot b_{7} \cdot b_{8} \cdot b_{9} \cdot \cdot \cdot \cdot b_{n} = \sum_{n}(log(log(b)-log(log(1)) + \cdot \cdot \cdot + (log(log(b)-log(log(1)) < \infty$$

Applying $(1.1)$ to our subtle but recent observations we know have the result:

$$b_{n}=\sum_{n}^{}(\log(\log(b)-\log(\log(n))=\int_{0}^{m}\log(\log(b)-\log(\log(n))+\theta(\log(\log(b)-log(log(n))n-(\log(\log(b)-log(log(n))1 < \infty$$

Which from brief observation the integral on the RHS side can be evaluated or approximated to a finite value which is obviously less then infinity.

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the logarithm is a continuous function for all $x>1$. Hence it commutes with limits in this domain. As each $b_n>1$, all $P_N = \prod_{n\le N}b_n>1$, and so

$$\lim\log (P_N) = \log(\lim P_N)$$

which is exactly your statement when you apply the functional equation for the logarithm function. And since it is monotone, if the $P_N\to\infty$, so too is $\sum \log(b_n)\to\infty$.

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  • $\begingroup$ Interesting I didn't take note of this at the of answering my question was my initial approach to cumbersome $\endgroup$ – Zophikel May 26 '17 at 16:48
  • $\begingroup$ @Zophikel Yes, definitely. All you need is some calculus and the definitions! :) $\endgroup$ – Adam Hughes May 26 '17 at 16:53

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