MHB Evaluate Finite Summation Expression

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The discussion focuses on evaluating the finite summation expression \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k}. Participants suggest starting by writing out the first and last terms to gain insight into the evaluation process. A potential method for finding an explicit expression involves using the Noerdlund-Rice Integral, which connects the sum to a complex integral. The integral formulation requires careful selection of the function and path, indicating a complex evaluation process. Overall, the conversation highlights the mathematical intricacies involved in solving the summation.
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How to evaluate the following expression?[math] \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k} [/math]
regards,
Bincy
 
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bincybn said:
How to evaluate the following expression?[math] \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k} [/math]
regards,
Bincy

Have you tried writing out the first few terms and the last 2 terms?
 
But I didn't get the ans.
 
bincybn said:
How to evaluate the following expression?[math] \sum_{i=0}^{N} \binom{N}{i} \left(-1\right)^{i}\left(\frac{1}{2+i}\right)^{k} [/math]

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

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

$\chi$ $\sigma$
 
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