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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]

Questions

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!!

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# Homework Help: Confused with a series problem

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