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 similar 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 existence 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.

 

[wais]

1. Yes, your solution is correct. You have correctly used the geometric series formula to find the upper bound for the remainder term in both cases.

2. The reason for using " \leq " instead of " < " is because the terms in the series are positive. If we use " < ", it would imply that the remainder term is strictly less than the upper bound, but since all terms are positive, it is possible for the remainder term to be equal to the upper bound. Therefore, we use " \leq " to include the possibility of equality.

3. It may seem counter-intuitive, but it is possible for an increasing sequence to still have a limit less than 1. For example, if the sequence is 1, 1/2, 3/4, 7/8, 15/16, ... then the limit is 1, but the sequence is still increasing. It is important to note that in the problem, we are given that the limit is less than 1, not specifically that it is decreasing. Therefore, we must consider both cases for the sequence.