<p>sum of (-1)((n^2)/(n^3+4)) as the sum goes from 1 to infinity.
why is this convergent?</p>
<p>help!</p>
<p>the problem is actually sum of (-1)^(n)((n^2)/(n^3+4)) as the sum goes from 1 to infinity.
why is this convergent?</p>
<p>help!</p>
<p>i assulme you have to use the alternating test, but im kind of confused on how to do so</p>
<p>Actually this is a very stupid problem, just take the limit as n goes to infinity, by the nth divergence test, the series goes to 0 as n approaches infinity, so it is convergent.</p>
<p>In most cases, they won’t give yuo questions where it won’t satisfy the conditions.</p>
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<p>Your post is entirely wrong. The nth-term divergence test CANNOT PROVE CONVERGENCE. EVER. 1/n approaches 0 as n approaches infinity, but the series diverges. In this case, you use the alternating series test. The series is of the form (-1)^n * s<em>n, so it is an alternating series, and because the limit of s</em>n as n approaches infinity is 0 and there exists an N such that if n > N, s<em>n > s</em>(n+1), the series converges.</p>
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<p>oh right, I don’t know why I put that there lol, but my Maths is correct.</p>
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LOL
look at how angry someone can get over Calculus. funny stuff.
But he is right.
the two things that satisfy the alternating series test are that the nth term test = 0 and that each consecutive term of the sequence is less than the previous one. That’s why it converges.</p>
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<p>I wasn’t trying to sound angry so much as I was trying to make it clear to the creator of this thread that the nth-term divergence test was not why the series converged, as he seemed to be confused about a relatively simple topic.</p>
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<p>There are actually three things; the series in question has to be an alternating series :p</p>
<p>wait… isn’t it the absolute value of the terms have to be decreasing?</p>
<p>that is what you check when you disregard the (-1)^n and see if s(n)>s(n+1)</p>
<p>Alternating series test. You remove the (-1)^n and look at what’s left. If it’s decreasing, and goes to 0 as n–>infinity, then the series converges. If not (if one or both criteria fail), then it diverges.</p>
<p>Pretty sure the above poster’s (314159265’s) comment about alternating series is too strong. It’s possible for the decreasing criteria to fail and yet for the alternating series to still converge.</p>
<p>Consider the series a<em>n from n = 1 to infinity, where a</em>n = -1/(n^2) when n is odd and a_n = 1/(n^3) when n is even. The terms of this series are:</p>
<p>-1 + 1/8 - 1/9 + 1/64 - 1/25 + 1/216 - 1/49 + …</p>
<p>This particular series converges despite failing the second criteria of the alternating series test.</p>
<p>For those still not convinced of the convergence of the series, you could rewrite the series as explained above as a_n = -1/(n^2) + 1/(n+1)^3 from n = 1 to infinity.</p>
<p>Then a<em>n = [-1(n+1)^3 + n^2]/[n^2(n+1)^3] = -1[(n+1)^3 - n^2]/[n^2(n+1)^3]. a</em>n has degree 3 in the numerator and degree 5 in the denominator. Performing a limit comparison test with the convergent p-series 1/(n^2) confirms the convergence of this series.</p>
<p>On the other hand, if the first criteria fails, the series does indeed always diverge by the n-th term Test, as has been pointed out by numerous posters.</p>
<p>But your problem would not really be approached as a “normal” alternate series would be approached. I’d make yours:</p>
<p>series a<em>n from n=0 to infinity, a</em>n = -1/(2n+1)^2 plus the series b<em>n from n=1 to infinity, b</em>n = 1/(2n)^3.</p>
<p>The second series converges because it is a p-series, and you can compare the first series to a p-series to find it also converges, so the complete series converges.</p>
<p>^ Agreed. I’m not saying that I would approach the series as a standard alternating series question either. I’m just saying that a series that is alternating that fails the alternating series test isn’t required to diverge.</p>
<p>MathProf, your series of a_n of -1/(n^2) + 1/(n+1)^3 is not alternating (at least by the terms of the alternating series test). You can split the terms into 2 summations. The first summation would be -1/(n^2), which converges by p. The second summation is 1/(n+1)^3, which I believe converges by integral test (check me on that). </p>
<p>I may have been a bit hasty by the terms about the alternating test, though. The whole a<em>n decreases as it goes to 0 is applicable after you disregard a (-1) to a certain power. After removing a (-1)^(…), you take what’s left and call that a</em>n. If it’s decreasing and going to 0 as n–>infinity (check with calculator), it converges.</p>
<p>You basically solved it like I did (see two posts above yours). However, his series IS and alternating series. The definition of an alternating series is one that alternates from + to - to + to - and so on and so forth.</p>
<p>[Alternating</a> series test - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Alternating_series_test]Alternating”>Alternating series test - Wikipedia)</p>
<p>It’s not an alternating series by this definition/series type. The series must be in this form for the test to be applicable. Of course, his series is alternating, but this test is inapplicable, which is why the series converges while not being monotone decreasing.</p>
<p>"Pretty sure the above poster’s (314159265’s) comment about alternating series is too strong. It’s possible for the decreasing criteria to fail and yet for the alternating series to still converge.</p>
<p>Consider the series a<em>n from n = 1 to infinity, where a</em>n = -1/(n^2) when n is odd and a_n = 1/(n^3) when n is even. The terms of this series are:</p>
<p>-1 + 1/8 - 1/9 + 1/64 - 1/25 + 1/216 - 1/49 + …</p>
<p>This particular series converges despite failing the second criteria of the alternating series test.</p>
<p>For those still not convinced of the convergence of the series, you could rewrite the series as explained above as a_n = -1/(n^2) + 1/(n+1)^3 from n = 1 to infinity.</p>
<p>Then a<em>n = [-1(n+1)^3 + n^2]/[n^2(n+1)^3] = -1[(n+1)^3 - n^2]/[n^2(n+1)^3]. a</em>n has degree 3 in the numerator and degree 5 in the denominator. Performing a limit comparison test with the convergent p-series 1/(n^2) confirms the convergence of this series.</p>
<p>On the other hand, if the first criteria fails, the series does indeed always diverge by the n-th term Test, as has been pointed out by numerous posters." </p>
<p>such a series would not appear on the AP test.</p>
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<p>Actually, it fits the Wikipedia definition perfectly if a_n is defined as 1/n^2, where n is odd, and 1/n^3, where n is even.</p>
<p>Also, the very first statement of the Wikipedia link states that the “The alternating series test is a method used to prove that infinite series of terms converge” (emphasis mine). You simply can’t say that a series diverges by the Alternating Series Test, similarly to how you can’t say that a series converges by the n-th Term Test.</p>
<p>And while I’m 99% sure that bobtheboy is right that a series would likely not appear on the AP Test, it also stands to reason that most of you have learned a few things during the year that won’t appear on the AP Test. :)</p>
<p>^^
I think what he meant to say (or should have said) is not that it wasn’t an alternating series, but that you can’t use the alternating series test for your series because it is not in the form</p>
<p>(-1)^n * a_n</p>
<p>which is required to use the alternating series test. But there’s no reason we need to explain your problem anymore since I’m pretty sure we all understand to approach it in a different way and that the alternating series test should be used only for the form above. And you never know - this could be on the AP exam this year!</p>
<p>The alternating series test can tell you converge or diverge (though absolute and conditional convergence needs to be determined with another test, if asked). MathProf, your series doesn’t fit the (-1)^…*a_n mold for the Alternating Series test. I wouldn’t be surprised if some series like that does show up on an AP exam (though it might be 2 geometric series rolled into one fraction; MathProf’s series is a bit too time-consuming (especially the 2nd part) to really be usable on an AP Exam).</p>