Confused with a series problem...

[DivGradCurl]

Problem

Let \sum a_n be a series with positive terms and let r_n = \frac{a_{n+1}}{a_n} . Suppose that \lim _{n \to \infty} r_n = L < 1 , so \Sum a_n converges by the Ratio Test. As usual, we let R_n be the remainder after n terms, that is,

R_n = a_{n+1} + a_{n+2} + a_{n+3} + \cdots

(a) If \left\{ r_n \right\} is a decreasing sequence and r_{n+1} < 1 , show by summing a geometric series, that

R_n \leq \frac{a_{n+1}}{1-r_{n+1}}

(b) If \left\{ r_n \right\} is an increasing sequence, show that

R_n \leq \frac{a_{n+1}}{1-L}

My Solution

(a) The first term r_{n+1} prevails, since it represents an upper bound to other terms of \left\{ r_n \right\} . Then,

R_n \leq a_{n+1} + a_{n+1}r_{n+1} + a_{n+1}\left( r_{n+1} \right) ^2 + a_{n+1}\left( r_{n+1} \right) ^3 + \cdots = \sum _{n=1} ^{\infty} a_{n+1}\left( r_{n+1} \right) ^{n-1} = \frac{a_{n+1}}{1-r_{n+1}}

(b) The last term L prevails, since it represents an upper bound to other terms of \left\{ r_n \right\} . Then,

R_n \leq a_{n+1} + a_{n+1}L + a_{n+1}L ^2 + a_{n+1} L ^3 + \cdots = \sum _{n=1} ^{\infty} a_{n+1}L ^{n-1} = \frac{a_{n+1}}{1-L}

Questions

1. Did I get it right?
2. Why use " \leq " instead of " < ", since all terms of \left\{ r_n \right\} are smaller then their respective upper bounds?
3. Isn't an increasing \left\{ r_n \right\} rather contra-intuitive when we consider a convergent series \sum a_n , which obeys: \lim _{n \to \infty} a_n =0 ?

That's it. Thank you very much!!

[NateTG]

Well, it really should be:
\lim_{n\rightarrow \infty} | r_n | < 1
otherwise, it would be possible to sneak in things like \sum 2^{-2n} where r_n=-2<1.

Regarding your answers:

If you can assume that |r_n| is decreasing and |r_{n+1}|<1
You need to use absolute values to make the inequalities work:
|\sum_{i=n+1}^\infty a_i| \leq \sum_{i=n+1}^\infty |a_i| \leq \sum_{i=n+1}^\infty |a_n r_{n+1}^{i-n-1}|=\sum_{i=1}^\infty |a_n| |r_{n+1}|^i
You have the right idea, but what you have is not true if r_{n+1} is positive and a_n is negative.

There is a similiar issue with your second answer.

Regarding the use of \leq rather than <:
Constant sequences are often included in the notion of decreasing or increasing sequence, so there may be equality rather than strict inequality.

Regarding the existance of increasing {r_n}
Consider, for example the possibility that r_n=\frac{1}{2}-\frac{1}{2^{n+1}}. It's pretty easy to see that the limit \lim_{n\rightarrow \infty} r_n = \frac{1}{2} and that \{r_n\} is increasing. However, since all of the r_n are positive and less than \frac{1}{2} we have:
0< r_n < \frac{1}{2}
Now
a_n=a_{n-1}\times r_{n-1}
so
\sum_{i=0}^{\infty} a_n = \sum_{i=0}^{\infty} (a_0 \times \prod_{j=0}^{i} r_j) < \sum_{i=0}^{\infty} (a_0 \times \prod_{j=0}^{i}\frac{1}{2}) = a_0 \times \sum_{i=0}^{\infty} \frac{1}{2^n} = 2a_0
which means that although \{r_n\} is increasing the sum is convergent.