A calculus problem by Rocco Dalto

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

∑ 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}} }

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Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Sep 9, 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}} }

1 Helpful 0 Interesting 0 Brilliant 0 Confused

What are b1, b2 , b3, ... , bk? Are they constants? How do we determine these values?

Pi Han Goh - 4 years, 9 months ago

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

Given ∑ 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} } } , you want to obtain the next summation ∑ N = 1 ∞ N K + 1 X N {\bf \sum_{N = 1}^{\infty} \frac{N^{K + 1}}{X^N} } in terms of these constants.

For instance, if you derived ∑ N = 1 ∞ N 3 X N = X 3 + 4 X 2 + X ( X − 1 ) 4 \sum_{N=1}^\infty \dfrac{N^3}{X^N} = \frac{X^3 + 4X^2 + X}{(X - 1)^4} ,
then using the next summation above we obtain: ∑ N = 1 ∞ N 4 X N \sum_{N=1}^\infty \frac{N^4}{X^N} = X 4 + 11 X 3 + 11 X 2 + X ( X − 1 ) 5 . = \frac{X^4 + 11X^3 + 11X^2 + X}{(X - 1)^5}.

Starting with ∑ N = 1 ∞ N X N \sum_{N = 1}^{\infty} \frac{N}{X^N} = = X ( X − 1 ) 2 \frac{X}{(X - 1)^2} we can obtain each next summation using the above.

Rocco Dalto - 4 years, 9 months ago

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