# Confused with a series problem

1. Oct 20, 2004

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!!

2. Oct 20, 2004

### 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 $$r_n=-2<1$$.

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.

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.

Last edited: Oct 20, 2004