Integrals and Summations

Calculus Level pending

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

I f If ∑ N = 1 ∞ N K X N {\bf \sum_{N = 1}^{\infty} \frac{N^K}{X^N} } = = ∑ J = 0 K − 1 b K − J ∗ X K − J ( X − 1 ) K + 1 {\bf \frac{\sum _{J = 0}^{K - 1} b_{K - J} * X^{K - J}}{(X - 1)^{K + 1} } }

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

t h e n then

(A:) ∑ N = 1 ∞ N K + 1 X N {\bf \sum_{N = 1}^{\infty} \frac{N^{K + 1}}{X^N} } = = X K + 1 + ( ∑ J = 0 K − 2 ( ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − J − 1 ) ∗ X K − J ) + X ( X − 1 ) K + 2 {\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}} } a n d and (B:) ∫ 2 3 ∑ N = 1 ∞ N K + 1 X N d x {\bf \int_{2}^{3} \sum_{N = 1}^{\infty} \frac{N^{K + 1}}{X^N} dx } =

ln ⁡ ( 2 ) + ( ∑ J = 1 K + 1 ( K + 1 ) ! ( K + 1 − J ) ! ∗ J ! ∗ J ∗ ( 1 − 1 2 J ) ) {\bf \ln(2) + (\sum_{J = 1}^{K + 1} \frac{(K + 1)!}{(K + 1 - J)! * J! * J} * (1 - \frac{1}{2^J}) )} + ( ∑ J = 0 K − 2 ( ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − J − 1 ) ∑ M = 0 K − J ( ( K − J ) ! K − J − M ) ! ∗ M ! ∗ ( 1 J + M + 1 ) ∗ ( 1 − 1 2 J + M + 1 ) ) {\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}})) } + ( 1 K ∗ ( 1 − 1 2 K ) + 1 K + 1 ∗ ( 1 − 1 2 K + 1 ) ) {\bf + (\frac{1}{K} * (1 - \frac{1}{2^K}) + \frac{1}{K + 1} * (1 - \frac{1}{2^{K + 1}} ))}

A and B are both true A and B are both false A is true and B is False A is False and B is True

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Sep 12, 2016

Let K {\bf K } be a positive integer.

Replace X X by 1 X \frac{1}{X} in ∑ N = 1 ∞ N K X N {\bf \sum_{N = 1}^{\infty} \frac{N^K}{X^N} } = = ∑ J = 0 K − 1 b K − J ∗ X K − J ( X − 1 ) K + 1 {\bf \frac{\sum _{J = 0}^{K - 1} b_{K - J} * X^{K - J}}{(X - 1)^{K + 1} } } ⟹ \implies

∑ N = 1 ∞ N K ∗ X N {\bf \sum_{N = 1}^{\infty} N^K * X^N } = = ∑ J = 0 K − 1 b K − J ∗ X J + 1 ( 1 − X ) K + 1 {\bf \frac{\sum_{J = 0}^{K - 1} b_{K - J} * X^{J + 1}}{(1 - X)^{K + 1}} }

d / d x ( ∑ N = 1 ∞ N K ∗ X N ) {\bf d/dx(\sum_{N = 1}^{\infty} N^K * X^N) } = = ∑ J = 0 K − 1 ( J + 1 ) ∗ b K − J ∗ X J − ∑ J = 0 K − 1 ( J + 1 ) ∗ b K − J ∗ X J + 1 + ( K + 1 ) ∗ ∑ J = 0 K − 1 b K − J ∗ X J + 1 ( 1 − X ) K + 2 {\bf \frac{\sum_{J = 0}^{K - 1} (J + 1) * b_{K - J} * X^J - \sum_{J = 0}^{K - 1} (J + 1) * b_{K - J} * X^{J + 1} + (K + 1) * \sum_{J = 0}^{K - 1} b_{K - J} * X^{J + 1} } {(1 - X)^{K + 2}} } = =

[ b K + ( K ∗ b K + 2 ∗ b K − 1 ) ∗ X + ( ( K − 1 ) ∗ b K − 1 + 3 ∗ b K − 2 ) ∗ X 2 + . . . + {\bf [ b_{K} + (K * b_{K} + 2 * b_{K - 1}) * X + ((K - 1) * b_{K - 1} + 3 * b_{K - 2}) * X^2 + ... + }

( ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − J − 1 ) ∗ X J + 1 + . . . + ( 2 ∗ b 2 + K ∗ b 1 ) ∗ X K − 1 + b 1 ∗ X K ] {\bf ((K - J) * b_{K - J} + (J + 2) * b_{K - J - 1}) * X^{J + 1} + ... + (2 * b_{2} + K * b_{1}) * X^{K - 1} + b_{1} * X^K ] } / ( 1 − X ) K + 2 {\bf /(1 - X)^{K + 2} } = = b K + ( ∑ J = 0 K − 2 ( ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − j − 1 ) ∗ X J + 1 ) + b 1 ∗ X K ( 1 − X ) K + 2 {\bf \frac{b_{K} + (\sum_{J = 0}^{K - 2} ((K - J) * b_{K - J} + (J + 2) * b_{K - j - 1}) * X^{J + 1}) + b_{1} * X^K}{(1 - X)^{K + 2}} } ⟹ \implies

∑ N = 1 ∞ N K + 1 ∗ X N {\bf \sum_{N = 1}^{\infty} N^{K + 1} * X^N } = = b K ∗ X + ( ∑ J = 0 K − 2 ( ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − j − 1 ) ∗ X J + 2 ) + b 1 ∗ X K + 1 ( 1 − X ) K + 2 {\bf \frac{b_{K} * X + (\sum_{J = 0}^{K - 2} ((K - J) * b_{K - J} + (J + 2) * b_{K - j - 1}) * X^{J + 2}) + b_{1} * X^{K + 1}}{(1 - X)^{K + 2}} }

Replacing X X by 1 X \frac{1}{X} ⟹ \implies

∑ N = 1 ∞ N K + 1 X N {\bf \sum_{N = 1}^{\infty} \frac{N^{K + 1}}{X^N} } = = X K + 1 + ( ∑ J = 0 K − 2 ( ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − J − 1 ) ∗ X K − J ) + X ( X − 1 ) K + 2 {\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}} }

N o w Now L e t Let U = X − 1 {\bf U = X - 1 } a n d and

I 1 = ∫ 1 2 ( U + 1 ) K + 1 U K + 2 d u {\bf I_{1} = \int_{1}^{2} \frac{(U + 1)^{K + 1}}{U^{K + 2}} du }

L e t Let a J = ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − J − 1 {\bf a_{J} = (K - J) * b_{K - J} + (J + 2) * b_{K - J - 1} } a n d {\bf and } I 2 = ∫ 1 2 ∑ J = 0 K − 2 a J ∗ ( U + 1 ) K − J U K + 2 d u {\bf I_{2} = \int_{1}^{2} \sum_{J = 0}^{K - 2} a_{J} * \frac{(U + 1)^{K - J}}{U^{K + 2}}du }

a n d and I 3 = ∫ 1 2 U + 1 U K + 2 {\bf I_{3} = \int_{1}^{2} \frac{U + 1}{U^{K + 2}} }

For I 1 : {\bf I_{1}: }

I 1 = ∫ 1 2 ( U + 1 ) K + 1 U K + 2 d u = {\bf I_{1} = \int_{1}^{2} \frac{(U + 1)^{K + 1}}{U^{K + 2}} du } = ∫ 1 2 1 U d u + ∑ J = 1 K + 1 ( K + 1 ) ! ( K + 1 − J ) ! ∗ J ! ∫ 1 2 U − ( J + 1 ) d u = {\bf \int_{1}^{2} \frac{1}{U} du + \sum_{J = 1}^{K + 1} \frac{(K + 1)!}{(K + 1 - J)! * J!} \int_{1}^{2} U^{-(J + 1)} du = }

ln ⁡ ( U ) − ∑ J = 1 K + 1 ( K + 1 ) ! ( K + 1 − J ) ! ∗ J ! ∗ J ∗ ( 1 U J ) ∣ 1 2 = {\bf \ln(U) - \sum_{J = 1}^{K + 1} \frac{(K + 1)!}{(K + 1 - J)! * J! * J} * (\frac{1}{U^J})|_{1}^{2} = } l n ( 2 ) + ∑ J = 1 K + 1 ( K + 1 ) ! ( K + 1 − J ) ! ∗ J ! ∗ J ∗ ( 1 − 1 2 J ) {\bf ln(2) + \sum_{J = 1}^{K + 1} \frac{(K + 1)!}{(K + 1 - J)! * J! * J} * (1 - \frac{1}{2^J}) }

∴ I 1 = l n ( 2 ) + ∑ J = 1 K + 1 ( K + 1 ) ! ( K + 1 − J ) ! ∗ J ! ∗ J ∗ ( 1 − 1 2 J ) {\bf \therefore I_{1} = ln(2) + \sum_{J = 1}^{K + 1} \frac{(K + 1)!}{(K + 1 - J)! * J! * J} * (1 - \frac{1}{2^J}) }

For I 2 : {\bf I_{2}: }

I 2 = ∫ 1 2 ∑ J = 0 K − 2 a J ∗ ( U + 1 ) K − J U K + 2 d u = {\bf I_{2} = \int_{1}^{2} \sum_{J = 0}^{K - 2} a_{J} * \frac{(U + 1)^{K - J}}{U^{K + 2}}du = }
∑ J = 0 K − 2 a J ∫ 1 2 ( U + 1 ) K − J U K + 2 d u = {\bf \sum_{J = 0}^{K - 2} a_{J} \int_{1}^{2} \frac{(U + 1)^{K - J}}{U^{K + 2}} du = } ∑ J = 0 K − 2 a J ∑ M = 0 K − J ( K − J ) ! ( K − J − M ) ! ∗ M ! ∫ 1 2 U − ( J + M + 2 ) d u = {\bf \sum_{J = 0}^{K - 2} a_{J} \sum_{M = 0}^{K - J} \frac{(K - J)!}{(K - J - M)! * M!} \int_{1}^{2} U^{-(J + M + 2)} du = } ∑ J = 0 K − 2 a J ∑ M = 0 K − J ( K − J ) ! ( K − J − M ) ! ∗ M ! ∗ ( − 1 J + M + 1 ) ∗ ( 1 U J + M + 1 ) ∣ 1 2 = {\bf \sum_{J = 0}^{K - 2} a_{J} \sum_{M = 0}^{K - J} \frac{(K - J)!}{(K - J - M)! * M!} * (\frac{-1}{J + M + 1}) * (\frac{1}{U^{J + M + 1}})|_{1}^{2} = }

∑ J = 0 K − 2 a J ∑ M = 0 K − J ( K − J ) ! ( K − J − M ) ! ∗ M ! ∗ ( 1 J + M + 1 ) ∗ ( 1 − 1 2 J + M + 1 ) {\bf \sum_{J = 0}^{K - 2} a_{J} \sum_{M = 0}^{K - J} \frac{(K - J)!}{(K - J - M)! * M!} * (\frac{1}{J + M + 1}) * (1 - \frac{1}{2^{J + M + 1}}) }

∴ I 2 = {\bf \therefore I_{2} = } ∑ J = 0 K − 2 ( ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − J − 1 ) ∑ M = 0 K − J ( K − J ) ! ( K − J − M ) ! ∗ M ! ∗ ( 1 J + M + 1 ) ∗ ( 1 − 1 2 J + M + 1 ) {\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}}) }

For I 3 : {\bf I_{3}: }

I 3 = ∫ 1 2 U − ( K + 1 ) + U − ( K + 2 ) d u = {\bf I_{3} = \int_{1}^{2} U^{-(K + 1)} + U^{-(K + 2)} du = } 1 K ∗ ( 1 − 1 2 K ) + 1 K + 1 ∗ ( 1 − 1 2 K + 1 ) {\bf \frac{1}{K} * (1 - \frac{1}{2^K}) + \frac{1}{K + 1} * (1 - \frac{1}{2^{K + 1}}) }

∴ \therefore ∫ 2 3 ∑ N = 1 ∞ N K + 1 X N d x {\bf \int_{2}^{3} \sum_{N = 1}^{\infty} \frac{N^{K + 1}}{X^N} dx } =

ln ⁡ ( 2 ) + ( ∑ J = 1 K + 1 ( K + 1 ) ! ( K + 1 − J ) ! ∗ J ! ∗ J ∗ ( 1 − 1 2 J ) ) {\bf \ln(2) + (\sum_{J = 1}^{K + 1} \frac{(K + 1)!}{(K + 1 - J)! * J! * J} * (1 - \frac{1}{2^J}) )} + ( ∑ J = 0 K − 2 ( ( K − J ) ∗ b K − J + ( J + 2 ) ∗ b K − J − 1 ) ∑ M = 0 K − J ( ( K − J ) ! K − J − M ) ! ∗ M ! ∗ ( 1 J + M + 1 ) ∗ {\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 − 1 2 J + M + 1 ) ) {\bf(1 - \frac{1}{2^{J + M + 1}})) } + ( 1 K ∗ ( 1 − 1 2 K ) + 1 K + 1 ∗ ( 1 − 1 2 K + 1 ) ) {\bf + (\frac{1}{K} * (1 - \frac{1}{2^K}) + \frac{1}{K + 1} * (1 - \frac{1}{2^{K + 1}} ))}

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