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Alternating and diverging series- use Leibniz test? Watch

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    I'm asked to find whether this sum (absolutely)Converges or Diverges \displaystyle\sum_{n=2}^ \infty b_n

    when \displaystyle b_n=(-1)^n\frac{(n+1)^n}{n^n}.

    I know it's an alternating series since because of (-1)^n and that \displaystyle\lim_{n \rightarrow \infty}b_n=e, which means the series diverges since e is obviously increasing.

    But how can I use Leibniz test(aka Alternating Series Test) to show this series is divergent when the Leibniz test is used to show a series converging?

    I could use the Divergence test since e doesn't equal {0}.
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    Can you remember the conditions that need to be met in order to show that a series converges using the leibniz theorem?
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    Surely for the Leibniz test to be applicable, you'd need  |a_{n+1}| < |a_{n}| where we define a_{n} = (\frac{n+1}{n})^n , which is clearly not true?
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    Hmm, I'm thinking of writing the same thing too...

    Since b_n \neq {0}, according to Leinbnitz test b_n is not converging, hen the series \displaystyle\sum_{n=2}^ \infty b_n is sdivergent and alternating series.
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      For \displaystyle \sum b_n to be convergent, we must have b_n \to 0. But this isn't the case (in particular, |b_n| -> e so b_n doesn't tend to 0), so the series does not converge. Leibniz's test is a test for convergence, it doesn't show divergence.
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      So is there any point of using the Leibnitz test whatsoever for this problem? We already know the limit goes to e so the series must be diverging...
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        (Original post by TheNihilist)
        So is there any point of using the Leibnitz test whatsoever for this problem? We already know the limit goes to e so the series must be diverging...
        Nope, no point. Also, |b_n| goes to e, b_n doesn't tend to anything.
       
       
       
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