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- Thread starter bincybn
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- Thread starter
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Have you tried writing out the first few terms and the last 2 terms?How to evaluate the following expression?

\(\displaystyle \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k} \)

regards,

Bincy

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- Feb 13, 2012

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The explicit expression [if it exists...] of the finite sum may be [probably...] found using the so called 'Noerdlund- Rice Integral'...How to evaluate the following expression?

\(\displaystyle \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k} \)

$\displaystyle \sum_{j=\alpha}^{n} \binom{n}{j}\ (-1)^{j} f(j)= (-1)^{n} \frac{n!}{2\ \pi\ i}\ \int_{\gamma} \frac{f(z)}{z\ (z-1)\ (z-2)...(z-n)}\ dz$ (1)

... setting $\alpha=0$, $\displaystyle f(z)=\frac{1}{(2+z)^{k}}$ and with proper choice of the path $\gamma$. The details are quite complex and require more work...

Kind regards

$\chi$ $\sigma$