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So I have this series

$$\sum_{j=2}^{\infty} \frac{-(2j-3)(j-3)}{(5j-8)(4j+1)}$$

I figured by reasoning that with the leading coefficients: -2J^2 / 20 j^2

the bottom would win out if the J went to infinity.

Also it is ratio of 1/10 so I thought the series would converge.

So why does this series diverge?

Can someone use the Cauchy Condensation, Comparison Test, or Ratio test.

I haven't started reviewing integral test or the other tests yet.

Calilaun
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  • I think they call it the "divergence test" or whatever, but the terms don't tend to 0, which is a clear necessary condition for convergence (via the Cauchy criterion) – user217285 Aug 22 '16 at 21:45
  • The terms as you have observed are nearly equal to $-1/10$ after some point. Do you think $(-1/10) + (-1/10) + (-1/10) + \cdots $ converges? Point number 2 is: revisit the ratio test - carefully. – zhw. Aug 22 '16 at 22:36
  • The ratio test refers to the ratio between successive terms, not the value of the individual terms themselves. The ratio between successive terms tends to unity as $j$ increases without bound, so the ratio test is inconclusive. (The series itself clearly is unbounded, since the individual terms do not go to $0$, as Nitin observed.) – Brian Tung Aug 22 '16 at 22:58

3 Answers3

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It diverges: for any convergent series, the general term tends to $0$. Here the general term tends to $-\dfrac1{10}$.

Pedro
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Bernard
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The expression $$ -\frac{(2j-3)(j-3)}{(5j-8)(4j+1)}=-\frac{(2-3/j)(1-3/j)}{(5-8/j)(4+1/j)}$$ has a limit of $-(2)(1)/(5)(4)=-1/10$ as $j\to \infty.$ Since the terms of the series do not tend to $0,$ the series has no limit.

Pedro
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DanielWainfleet
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I see the flaw in my reasoning. I have done the ratio test and cauchy condesation test and found that it diverges.

Calilaun
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