Remainder bounds for series with ratio test conditions

[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.