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I would like to approximate $$ \ln(\sum_{k=0}^n(n-2k)^p) $$ Here $p\geq 2$

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up vote 1 down vote accepted

Presumably the upper limit should be $n/2$ to avoid powers of negative numbers (particularly problematic if $p$ isn't an odd integer). I would convert to an integral: $\sum_{k=0}^n(n-2k)^p \approx \int_{k=0}^{n/2}(n-2k)^p\;dk$ which yields to the substitution $u=n-2k$.

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Where did the logarithm go? – Tomarinator Apr 29 '12 at 7:35
@SauravTomar: Once the integral is evaluated, one can take the log. – Ross Millikan Apr 29 '12 at 15:50

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