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Limit proofs(indeterminate forms?)

  • #1
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limit proofs(indeterminate forms??)

Homework Statement



We work in the real numbers. Are the following true or false? Give a proof or counterexample.
(a) If [itex]\sum a^4_n[/itex] converges, then [itex]\sum a^5_n[/itex] converges.
(b) If [itex]\sum a^5_n[/itex] converges, then [itex]\sum a^6_n[/itex] converges.
(c) If [itex]a_n \geq 0[/itex] for all [itex]n[/itex], and [itex]\sum a_n[/itex] converges, then [itex]na_n \rightarrow 0[/itex] as [itex]n \rightarrow \infty[/itex].
(d) If [itex]a_n \geq 0[/itex], for all [itex]n[/itex], and [itex]\sum a_n[/itex] converges, then [itex]n(a_n - a_{n-1}) \rightarrow 0[/itex] as [itex]n \rightarrow \infty[/itex].
(e) If [itex]a_n[/itex] is a decreasing sequence of positive numbers, and [itex]\sum a_n[/itex] converges, then [itex]na_n \rightarrow 0[/itex] as [itex]n \rightarrow \infty[/itex].


Homework Equations





The Attempt at a Solution




(a) and (b) can be proved similarly. Since [itex]\sum a_n^4[/itex] converges, for some [itex]N[/itex], when [itex]n \geq N[/itex], then [itex]a_n^4 < 1[/itex]. Take [itex]\beta[/itex] s.t., [itex]a_n^4 < \beta < 1[/itex]. That is, [itex]a_n < (\beta)^{1/4} < 1[/itex]. Also, [itex]|(\beta)^{1/5}| < 1[/itex]. This implies that [itex]|a_n^5| < 1[/itex] and therefore [itex]\sum |a_n^5|[/itex] converges.

(c) This is where I get confused. THis seems like an indeterminate form, we have never done this in class. same for (d) and (e). any suggestions?
 

Answers and Replies

  • #2
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Your proof of (a) doesn't make sense. In order for a series [tex]\textstyle \sum b_n[/tex] to converge, it is not enough that [tex]|b_n| < 1[/tex] for sufficiently large [tex]n[/tex]!

Are you thinking of the root test? If so, there is certainly an exponent missing. But the converse of the root test is false: there are convergent series [tex]\textstyle\sum b_n[/tex] having [tex]|b_n|^{1/n} \to 1[/tex] as [tex]n\to\infty[/tex]. So you can't conclude from the convergence of [tex]\textstyle\sum a_n^4[/tex] anything about the behavior of [tex]|a_n^4|^{1/n}[/tex]. The root test is not useful here.

For (a) and (b) you should be thinking about comparisons. Bear in mind that you are not given that [tex]a_n \geq 0[/tex], so the proof that works for one of (a) and (b) will not necessarily work for the other.

For (c)-(e): The condition [tex]na_n \to 0[/tex] should make you think of a particular series whose behavior you know. Try to modify this series to construct examples that test the boundaries of the conditions here.
 
  • #3
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How does a series converge if its terms do not converge to 0?
 
  • #5
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so does my answer work? or no?
 
  • #6
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Are you thinking of the root test? If so, there is certainly an exponent missing. But the converse of the root test is false: there are convergent series [tex]\textstyle\sum b_n[/tex] having [tex]|b_n|^{1/n} \to 1[/tex] as [tex]n\to\infty[/tex]. So you can't conclude from the convergence of [tex]\textstyle\sum a_n^4[/tex] anything about the behavior of [tex]|a_n^4|^{1/n}[/tex]. The root test is not useful here.
I think you might be mistaken here. Did you mean that there are sequences that where [itex]a_n \rightarrow 0[/itex] but [itex]\sum a_n[/itex] doesn't converge? such as [itex]\sum \frac{1}{n}[/itex]
 
  • #7
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I think you might be mistaken here. Did you mean that there are sequences that where [itex]a_n \rightarrow 0[/itex] but [itex]\sum a_n[/itex] doesn't converge? such as [itex]\sum \frac{1}{n}[/itex]
No, I meant precisely what I said: there are convergent series whose convergence is not detected by the root test, so the converse of the root test is false. [tex]\sum\frac1{n^2}[/tex] is one; it converges by comparison to the telescoping series [tex]\sum\frac1{n(n+1)}[/tex], but [tex]|n^{-2}|^{1/n} \to 1[/tex] as [tex]n\to\infty[/tex].
 
  • #8
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so does my answer work? or no?
No. As I said, your argument for (a) establishes only that [tex]|a_n^5| < 1[/tex] for all sufficiently large [tex]n[/tex], which is very far from sufficient for convergence.
 

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