# Integrals and Summations

Calculus Level pending

Let $${\bf K }$$ be a positive integer and $${\bf X > 1}$$

$$If$$ $${\bf \sum_{N = 1}^{\infty} \frac{N^K}{X^N} }$$ $$=$$ $${\bf \frac{\sum _{J = 0}^{K - 1} b_{K - J} * X^{K - J}}{(X - 1)^{K + 1} } }$$

where $${\bf b_{1} = b_{K} = 1}$$ and each constant $$\bf b_{i}$$ where $${\bf (2 <= i <= K - 1) }$$ is a positive integer

$$then$$

(A:) $${\bf \sum_{N = 1}^{\infty} \frac{N^{K + 1}}{X^N} }$$ $$=$$ $${\bf \frac{X^{K + 1} + (\sum_{J = 0}^{K - 2} ((K - J) * b_{K - J} + (J + 2) * b_{K - J - 1}) * X^{K - J}) + X}{(X - 1)^{K + 2}} }$$ $and$ (B:) $${\bf \int_{2}^{3} \sum_{N = 1}^{\infty} \frac{N^{K + 1}}{X^N} dx }$$ =

$${\bf \ln(2) + (\sum_{J = 1}^{K + 1} \frac{(K + 1)!}{(K + 1 - J)! * J! * J} * (1 - \frac{1}{2^J}) )}$$ $${\bf + (\sum_{J = 0}^{K - 2} ((K - J) * b_{K - J} + (J + 2) * b_{K - J - 1}) \sum_{M = 0}^{K - J} (\frac{(K -J)!}{K - J - M)! * M!} * (\frac{1}{J + M + 1}) * (1 - \frac{1}{2^{J + M + 1}})) }$$ $${\bf + (\frac{1}{K} * (1 - \frac{1}{2^K}) + \frac{1}{K + 1} * (1 - \frac{1}{2^{K + 1}} ))}$$

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