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Exercise 3.12 Suppose $a_n>0$ and $\sum a_n$ converges. Put $$r_n=\sum_{m=n}^\infty a_m.$$

a) Prove that $$\frac{a_m}{r_m}+\cdots+\frac{a_n}{r_n}>1-\frac{r_n}{r_m},$$ if $m>n$ and deduce that $\sum\frac{a_n}{r_n}$ diverges.

If we want $>$, then each denominator on the LHS should be replace with the largest denominator. If $m>n$, then $r_m<r_n$. So shouldn't the RHS be

$$1-\frac{r_m}{r_n}?$$

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  • $\begingroup$ The left hand side of the displayed equation (the sum of fractions) would normally be written with increasing indices, suggesting that what's intended is $"if $n > m$, and deduce that..." So maybe there is a misprint. $\endgroup$ – John Hughes Jan 10 '15 at 3:24
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According to my edition of Rudin's Principles of Mathematical Analysis (third edition) This is the exercise in question.

enter image description here

In my copy it reads $m<n$ but you typed it as $m>n$ above, hence the confusion.

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  • $\begingroup$ Pulled out my copy of baby Rudin, was about to submit the same! $\endgroup$ – Clark Zinzow Jan 10 '15 at 3:33
  • $\begingroup$ Yes. Quite right. Thank you. :) $\endgroup$ – Tim Raczkowski Jan 10 '15 at 3:33
  • $\begingroup$ @TimRaczkowski As for proof technique, as I expect you noticed, it helps to replace all of the denominators to the largest denominator (in this case $r_m$) and then notice that the top $(a_m+\dots+a_n)$ resembles a difference of two $r$ terms. $\endgroup$ – JMoravitz Jan 10 '15 at 3:36
  • $\begingroup$ Yes. I'm not having a problem with the proof. Just the notation. :) $\endgroup$ – Tim Raczkowski Jan 10 '15 at 3:38

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