Ytukyg

$(a_n)$ is a sequence of strictly increasing positive numbers and has a lower as well as an upper bound. The question is, if the partial sum:

begin{align*}sumlimits_{n=1}^{k}left(frac{a_{n+1}}{a_n}-1right)&=left(frac{a_{1+1}}{a_1}-1right)+left(frac{a_{2+1}}{a_2}-1right)+cdots +left(frac{a_{k+1}}{a_k}-1right)\&=frac{a_2}{a_1}+frac{a_3}{a_2}+cdots +frac{a_{k+1}}{a_k}-k.end{align*}

converges? Because $(a_n)$ is a sequence of strictly increasing positive numbers, we have $a_{n+1}>a_n$ for all $n$. Because $(a_n)$ is bounded, we have an upper bound, e.g. $a$, and a lower bound, e.g. $b$, such that $b leq a_n leq a$ for all $n$. This means that $b leq a_n < a_{n+1} leq a$...

The problem with that is: if $k rightarrow infty$, then the series diverges, because of the $-k$ at the end of the expression?! Is there some kind of trick or something to use here?!

Thanks for your help in advance.

Best Regards,

Ahmed Hossam

We are given

1. 
   $(a_n)$ is a strictly increasing sequence


2. of positive terms


3. bounded from above.

Let's say $M$ is an upper bound. (*1) and (*2) implies for all $n$,

$$a_{n+1} > a_n > 0 implies frac{a_{n+1}}{a_n} > 1 iff frac{a_{n+1}}{a_n} - 1 > 0$$

As a sequence indexed by $k$, the partial sums $s_k stackrel{def}{=}sumlimits_{n=1}^k left(frac{a_{n+1}}{a_n} - 1right)$ is a sum of positive terms, so it is strictly increasing. More precisely, we have

$$
s_{k+1} - s_{k} =
sumlimits_{n=1}^{k+1} left(frac{a_{n+1}}{a_n} - 1right)
-
sumlimits_{n=1}^{k} left(frac{a_{n+1}}{a_n} - 1right)
= frac{a_{k+2}}{a_{k+1}} - 1 > 0
implies s_{k+1} > s_{k}
$$

By (*1) and (*2) again, we have $a_{n+1} > a_n ge a_1 > 0$. This implies
$$s_k = sum_{n=1}^k left(frac{a_{n+1}}{a_n} - 1 right)
= sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_n}
le sum_{n=1}^k frac{a_{n+1} - a_{n}}{a_1}
= frac{1}{a_1}sum_{n=1}^k (a_{n+1} - a_n)
$$
The last sum is a telescoping sum. Together with (*3), we obtain
$$s_k le frac{1}{a_1}(a_{k+1} - a_1) le frac{M-1}{a_1}$$
This means as a sequence, the partial sums is bounded from above too. Being increasing and bounded from above, the sequence $s_k$ converges. By definition, so does the series
$sumlimits_{n=1}^infty left(frac{a_{n+1}}{a_n} - 1right)$.

You can check that it is equivalent to determine whether the series of $logleft(frac{a_{n+1}}{a_n}right)$ converges, because $frac{a_{n+1}}{a_n} rightarrow 1$.

Edit: to be more clear:

$frac{a_{n+1}}{a_n}-1 sim logleft(frac{a_{n+1}}{a_n}right) > 0$ as $n$ goes to infinity.

Thus, the series $sum_n{frac{a_{n+1}}{a_n}-1}$ is convergent iff the series $sum_n{logleft(frac{a_{n+1}}{a_n}right)}$ is convergent.