2
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.