# Thread: Divergence of a series

1. ## Divergence of a series

Here's a challenge problem I just made up:

Let $\displaystyle \{a_n\}$ be a sequence of positive real numbers such that $\displaystyle \sum a_n = \infty$. Show that

$\displaystyle \frac{a_1}{a_0}+\frac{a_2}{a_0+a_1}+\frac{a_3}{a_0 +a_1+a_2}+ \dots = \infty$.

2. Let $\displaystyle S_n = \sum_{k=0}^{n}a_k$.

We want to show that $\displaystyle \sum_{k=0}^n\frac{\Delta S_k}{S_k}}$ diverges as $\displaystyle n\to \infty$.
( I chose this notation to suggest the similarity with $\displaystyle \int_{ [ a,\infty)} \tfrac{f'(x)}{f(x)}dx = +\infty$ , if $\displaystyle \displaystyle\lim_{x\to\infty} f(x) = +\infty$ )

We assume that $\displaystyle \frac{\Delta S_k}{S_k}} \to 0$ otherwise the series would diverge trivially.

Let $\displaystyle L_k = \log S_k$, then $\displaystyle \Delta L_k = \log S_{k+1} - \log S_k = \log \left( 1 +\frac{\Delta S_k}{S_k} \right) =\frac{\Delta S_k}{S_k} + O\left( \left(\frac{\Delta S_k}{S_k}\right)^2 \right)$

We observe that $\displaystyle \sum_{k=0}^n \Delta L_k = \log S_{n+1} - \log S_0$ diverges since $\displaystyle \log S_n \to \infty$ and now by the limit comparison test - both differences are always positive-, we get that $\displaystyle \sum_{k=0}^n\frac{\Delta S_k}{S_k}}$ diverges too.

Notation: $\displaystyle \Delta a_n = a_{n+1}-a_n$

3. Very nice Paul!

Here's my solution:

It's well known that if $\displaystyle x_n>0$ and $\displaystyle \sum {x_n}$ converges , then $\displaystyle \prod (1+x_n)$ converges also. Let $\displaystyle A_n = \sum_{k=0}^n a_k$. We have

$\displaystyle \prod_{k=1}^n (1+\frac{a_k}{A_{k-1}}) = \prod_{k=1}^n\frac{A_k}{A_{k-1}} = A_n/a_0$. Now since $\displaystyle A_n \to \infty$, we see that $\displaystyle \prod (1+\frac{a_k}{A_{k-1}})$ diverges, which implies that $\displaystyle \sum \frac{a_k}{A_{k-1}}$ diverges.

4. Originally Posted by Bruno J.
Very nice Paul!

Here's my solution:

It's well known that if $\displaystyle x_n>0$ and $\displaystyle \sum {x_n}$ converges , then $\displaystyle \prod (1+x_n)$ converges also. Let $\displaystyle A_n = \sum_{k=0}^n a_k$. We have

$\displaystyle \prod_{k=1}^n (1+\frac{a_k}{A_{k-1}}) = \prod_{k=1}^n\frac{A_k}{A_{k-1}} = A_n/a_0$. Now since $\displaystyle A_n \to \infty$, we see that $\displaystyle \prod (1+\frac{a_k}{A_{k-1}})$ diverges, which implies that $\displaystyle \sum \frac{a_k}{A_{k-1}}$ diverges.
Your solution is same as what came to my mind when I saw the question.
For those who might want to see a reference for the relation between the sum and the product used by Bruno J, please see [Theorem 7.4.6, 1].
I quote the result for convenience.
Lemma. Let $\displaystyle \{x_{n}\}_{n\in\mathbb{N}}$ be a sequence of nonnegative numbers, then
$\displaystyle \sum_{n\in\mathbb{N}}x_{n}$ and $\displaystyle \prod_{n\in\mathbb{N}}(1+x_{n})$ converge or diverge together.

References
[1] L.S. Hahn and B. Epstein, Classical Complex Analysis, Jones and Bartlett Publishers, Inc., London, 1996.