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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