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Confused with a series problem

  1. Oct 20, 2004 #1

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

    [tex] R_n = a_{n+1} + a_{n+2} + a_{n+3} + \cdots [/tex] ​

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

    [tex] R_n \leq \frac{a_{n+1}}{1-r_{n+1}} [/tex]​

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

    [tex] R_n \leq \frac{a_{n+1}}{1-L} [/tex]​

    My Solution

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

    [tex] 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}} [/tex]

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

    [tex] 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} [/tex]


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

    That's it. Thank you very much!!
  2. jcsd
  3. Oct 20, 2004 #2


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    Science Advisor
    Homework Helper

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

    Regarding your answers:

    If you can assume that [tex]|r_n|[/tex] is decreasing and [tex]|r_{n+1}|<1[/tex]
    You need to use absolute values to make the inequalities work:
    [tex]|\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[/tex]
    You have the right idea, but what you have is not true if [tex]r_{n+1}[/tex] is positive and [tex]a_n[/tex] is negative.

    There is a similiar issue with your second answer.

    Regarding the use of [tex]\leq[/tex] rather than [tex]<[/tex]:
    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 [tex]{r_n}[/tex]
    Consider, for example the possibility that [tex]r_n=\frac{1}{2}-\frac{1}{2^{n+1}}[/tex]. It's pretty easy to see that the limit [tex]\lim_{n\rightarrow \infty} r_n = \frac{1}{2}[/tex] and that [tex]\{r_n\}[/tex] is increasing. However, since all of the [tex]r_n[/tex] are positive and less than [tex]\frac{1}{2}[/tex] we have:
    [tex]0< r_n < \frac{1}{2}[/tex]
    [tex]a_n=a_{n-1}\times r_{n-1}[/tex]
    [tex]\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[/tex]
    which means that although [tex]\{r_n\}[/tex] is increasing the sum is convergent.
    Last edited: Oct 20, 2004
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