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Prove/disprove: We have a positive sequence $a_n$ such that $\displaystyle\lim_{n\to\infty}a_n=0$ so the series $\displaystyle\sum^\infty_{n=1} (-1)^na_n $ converge.

It's very similar to Leibniz alternating sum test but I think the statement is false so I'm trying to find a counter example. If there's a positive sequence that tends to zero but isn't monotone decreasing then the alternating sum test won't work, thus the series does not converge.

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The alternating series test only gives a suficient condition for convergence. –  Zircht Apr 29 at 16:31
2  
Take, e.g., $a_n=1/n^2$, $n$ odd; $a_n=1/n$, $n$ even. –  David Mitra Apr 29 at 16:33
    
@ᛥᛥᛥ you mean there can be an alternating series that doesn't satisfy the alternating sum test but still converge ? –  GinKin Apr 29 at 16:34
    
@GinKin Yes. Try to find a convergent series similar to the one given by David Mitra. –  Zircht Apr 29 at 16:40
    
@ᛥᛥᛥ $a_n=1/n^2$, $n$ is odd, $a_n=0$, $n$ is even. I see. –  GinKin Apr 29 at 16:42

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