It is asked if it is convergent or Divergent series
1/(1*2) + 2/(3*4) + 3/(5*6)... infinity
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Martin Sleziak
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What makes you say this has anything to do with a hyperbola? Did you mean to say parabola instead? If so, that is only the denominator and you neglected to take the numerator into account as well. – JMoravitz Jul 12 '19 at 17:32
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Really sorry, I posted wrong general term, this is the full que – Rohit Jain Jul 12 '19 at 17:40
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Note that in this new version you display no effort whatsoever to solve the question. – José Carlos Santos Jul 12 '19 at 17:52
1 Answers
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No; it diverges. First of all, note that $\dfrac n{n(n+1)}=\dfrac1{n+1}$. On the other hand, the series $\sum_{n=0}^\infty\frac1{n+1}$ is the harmonic series, which diverges.
The series that you wrote after editing your question is $\displaystyle\sum_{n=1}^\infty\frac n{(2n-1)2n}=\sum_{n=1}^\infty\frac1{4n-2}$ which diverges too, since $(\forall n\in\mathbb N):\dfrac1{4n-2}\geqslant\dfrac14\times\dfrac1n$ and since the harmonic series diverges.
José Carlos Santos
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"the last term tends to zero and the graph is of hyperbola so it must be converging" is certainly NOT true! (Actually there is no "last term". I assume you mean the limit as n goes to 0. Also the graph of n/n(n+1) is a sequence of individual points, not a hyperbola. I assume you mean the graph of y= x/(x(x+ 1)), changing the discrete variable, n, to the continuous variable, x.) The sum $\Sigma \frac{1}{n}$ has those properties but does not converge. – user247327 Jul 12 '19 at 17:17
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My guess is that you meant to comment the question, not my answer. – José Carlos Santos Jul 12 '19 at 17:21
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Really sorry, I posted wrong general term, this is the full question. – Rohit Jain Jul 12 '19 at 17:41