# If $a_n$ is a strictly increasing unbounded sequence, does $\sum_n \frac{a_{n+1} - a_n}{a_n}$ diverge?

So I've been thinking through some test cases. If $a_n = n$ then $\sum_n \frac{a_{n+1} - a_n}{a_n}$ is the harmonic series which diverges. And if $a_n = \sum_{k=1}^n 1/k$ then $\sum_n \frac{a_{n+1} - a_n}{a_n}$ diverges like $\sum_n 1/(n \log n)$. So that got me thinking, if $a_n$ is a strictly increasing unbounded sequence, does $\sum_n \frac{a_{n+1} - a_n}{a_n}$ necessarily diverge?

Suppose that $\frac{a_{n+1}-a_n}{a_n}=b_n$, and $\sum\limits_{n=1}^\infty b_n<\infty$.

Then $$1+b_n=\frac{a_{n+1}}{a_n}$$ hence, $$a_{n+1}=\prod_{k=1}^n (1+b_n)$$ Transforming the product into the exp of a sum and taking limits, $$\lim_{n\to\infty} a_n=\exp\left(\sum_{n=1}^\infty\ln(1+b_n)\right)<\exp\sum_{n=1}^\infty b_n<\infty$$

So if $a_n$ is not bounded, your sum must diverge.

Yes.

Let $b_n=\frac{a_{n+1}-a_n}{a_n}> 0$. Suppose to the contrary that $\sum_n b_n<A$. We show that $$a_{n+1}=a_n (1+b_n)=a_1\Pi_{1\le k\le n}(1+b_k)$$ converges, contradicting the premise.

In fact $$\Pi_{1\le k\le n}(1+b_k)\le (\frac{\sum 1+b_k}{n})^n\le (1+\frac{A}{n})^n\to e^A.$$

Yes!

$$\sum_{k=1}^p \frac{a_{n+k+1} - a_{n+k}}{a_{n+k}}\geq\sum_{k=1}^p\frac{a_{n+k+1} - a_{n+k}}{a_{n+p+1}} =\frac{a_{n+p+1}-a_{n+1}}{a_{n+p+1}} = 1-\frac{a_{n+1}}{a_{n+p+1}}$$

Notice $\lim\limits_{p\to\infty}\frac{a_{n+1}}{a_{n+p+1}}=0$, exists $N$, $\forall p\gt N$, such that

$$\sum_{k=1}^p \frac{a_{n+k+1} - a_{n+k}}{a_{n+k}}\gt \frac12$$

Ok, another proof :)

Let $U_n$ be : $U_n = \frac{A_n- A_{n-1}}{A_n} = 1 - \frac{A_{n-1}}{A_n}$

Then : $ln(A_n) -ln(A_{n-1}) = -ln(1-U_n)$

Suppose the series of general term $(U_n)$ converges:

$(U_n)$ -->0

=> $ln(A_n) - ln(A_{n-1}) = U_n + o(U_n)$

It means that $( ln(A_n) -ln(A_{n-1}) )$ is the term of a convergent series, which is absurd since $(A_n)$ diverges to the infinity.

Hence the series of general term $(U_n)$ diverges, always.