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Convergence Test

  • Thread starter rocomath
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  • #1
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[tex]\sum_{n=1}^{\infty}\frac{(-1)^{n}n}{n^{2}+25}[/tex]

Ratio Test

[tex]\lim_{n\rightarrow\infty}|\frac{(-1)^{n+1}(n+1)(n^{2}+25)}{[(n+1)^{2}+25](-1)^{n}n}|[/tex]

[tex]\lim_{n\rightarrow\infty}|\frac{n^{3}+n^{2}+25n+25}{ n^{3}+2n^{2}+26n}|=1[/tex]

Thus, the Ratio Test is inconclusive. So what should my next step be, or other suggestions? Hmm ...

Thanks!
 

Answers and Replies

  • #2
Dick
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How about an alternating series check?
 
  • #3
HallsofIvy
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In fact, since that is NOT a series of positive numbers, the ratio test doesn't apply anyway!
 
  • #4
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In fact, since that is NOT a series of positive numbers, the ratio test doesn't apply anyway!
Just because contains non-positive terms does not mean one cannot apply the ratio test, since it compares the ratio of the absolute value of [tex]a_{n+1}[/tex] and [tex]a_n[/tex]. His first attempt is fine; he just happened to have a series for which the ratio test is inconclusive.
 
  • #5
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Just because contains non-positive terms does not mean one cannot apply the ratio test, since it compares the ratio of the absolute value of [tex]a_{n+1}[/tex] and [tex]a_n[/tex]. His first attempt is fine; he just happened to have a series for which the ratio test is inconclusive.

Then can u show us a proof that shows that the ratio test is consistent and applies even when a series contains non-positive terms??
 
  • #6
morphism
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Then can u show us a proof that shows that the ratio test is consistent and applies even when a series contains non-positive terms??
I suppose you believe the ratio test holds for series whose terms are nonnegative. Suppose then we apply it to the series [itex]\sum |a_n|[/itex]: If [itex]\lim |a_{n+1}|/|a_n| < 1[/itex], then [itex]\sum |a_n|[/itex] converges. But this in turn implies that [itex]\sum a_n[/itex] converges. This follows from the completeness of the real numbers, i.e. that every cauchy sequence of reals convereges.

To see this, let [itex]S_n = a_1 + a_2 + ... + a_n[/itex]. Then for [itex]n \geq m[/itex],
[tex]|S_n - S_m| = |a_{m+1} + ... + a_n| \leq |a_{m+1}| + ... + |a_n|[/tex]

If [itex]\sum |a_n|[/itex] converges, we can make the term on the right as small as we want. So [itex](S_n)_n[/itex] is cauchy, and consequently [itex]\sum a_n = \lim S_n < \infty[/itex].
 
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  • #7
HallsofIvy
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I agree that what I said at first was misleading, possibly just completely wrong!

Of course, if the ratio test, applied to |an| showed that it converged, that would show that the series is absolutely convergent which immediately implies that the series is convergent.

If the ratio test does not work, if the limit of the ratio is 1 or even greater than 1, it is still possible that the original series converges. As Dick said originally, it is far better to apply the "alternating series test" here. If |an| is decreasing, then the series converges.
 
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  • #8
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I suppose you believe the ratio test holds for series whose terms are nonnegative. Suppose then we apply it to the series [itex]\sum |a_n|[/itex]: If [itex]\lim |a_{n+1}|/|a_n| < 1[/itex], then [itex]\sum |a_n|[/itex] converges. But this in turn implies that [itex]\sum a_n[/itex] converges. This follows from the completeness of the real numbers, i.e. that every cauchy sequence of reals convereges.

To see this, let [itex]S_n = a_1 + a_2 + ... + a_n[/itex]. Then for [itex]n \geq m[/itex],
[tex]|S_n - S_m| = |a_{m+1} + ... + a_n| \leq |a_{m+1}| + ... + |a_n|[/tex]

If [itex]\sum |a_n|[/itex] converges, we can make the term on the right as small as we want. So [itex](S_n)_n[/itex] is cauchy, and consequently [itex]\sum a_n = \lim S_n < \infty[/itex].
OOh yeah, i forgot all about that. I just did not reflect on this at all. I've got to whatch my mouth next time.
 

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