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Feb 4 |
accepted | Sum involving binomial coefficients |
Feb 4 |
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Sum involving binomial coefficients
Thanks for the answer NS! |
Feb 4 |
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Sum involving binomial coefficients
Oh I did a mistake in my evaluation I think. Interesting, so it is just the harmonic series. |
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Sum involving binomial coefficients
But when I evalute that integral and the Stirling formula you gave, I don't get the same answer. How do you find that formula? |
Feb 4 |
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Sum involving binomial coefficients
I think one has to involve Stirling functions of the second kind, not Stirling numbers of the first kind. |
Feb 4 |
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Sum involving binomial coefficients
Are you sure about this? Then this would just reduce to $\sum_{j=1}^k 1/k$, right? |
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awarded | Scholar |
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awarded | Student |
Feb 4 |
asked | Sum involving binomial coefficients |