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I have an alternating sum and am looking for its asymptotic behaviour. Due to the change of signs there occure a lot of cancellations between the different terms. However the terms are all different, so no terms cancel exactly. Is there still a method that can be used to find the dominating behaviour of the sum?

Thanks

user404302
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  • https://en.wikipedia.org/wiki/Euler_summation might help, but this question seems too general. Can you post the summation? – adfriedman Mar 26 '17 at 07:12

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You can group pairs of terms together and study the new sum.

For example, if your series is $\sum_{n=1}^{\infty}(-1)^n/n$, you can group the $(2n)^{th}$ and $(2n+1)^{th}$ terms together. They nearly cancel out to give $1/(2n)(2n+1)$, so $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{n}=1+\sum_{n=1}^{\infty}\frac{1}{2n(2n+1)}, $$ and so the odd partial sums of the first series have the same asymptotic behavior as the partial sums on the right, whereas the even partial sums have a (slightly different) asymptotic behavior.

pre-kidney
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