Integral of a summation

Calculus Level 4

Evaluate ∫ 2 3 ∑ N = 1 ∞ N 3 x N d x \displaystyle \int_2^3 \sum_{N=1}^\infty \dfrac{N^3}{x^N} \, dx .

Use ln ⁡ 2 = 0.693147 \ln 2 = 0.693147 .

Give your answer to 6 decimal places.


The answer is 10.443147.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Sep 8, 2016

B e g i n Begin

L e t Let X > 1. X > 1. ∑ N = 1 ∞ N X N \sum_{N = 1}^{\infty} \frac{N}{X^N} = 1 X ∗ \frac{1}{X} * ( 1 + 1 X (1 + \frac{1}{X} + 1 X 2 \frac{1}{X^2} + . . . ) + + ...) + 1 X 2 ∗ \frac{1}{X^2} * ( 1 + 1 X (1 + \frac{1}{X} + 1 X 2 \frac{1}{X^2} + . . ) + . . . = + ..) + ... = X X − 1 \frac{X}{X - 1} * 1 X ∗ \frac{1}{X} * X X − 1 = \frac{X}{X - 1} = X ( X − 1 ) 2 \frac{X}{(X - 1)^2}

∑ N = 1 ∞ N 2 X N = \sum_{N = 1}^{\infty} \frac{N^2}{X^N} = ∑ J = 1 ∞ ∑ K = J ∞ K X K = \sum_{J = 1}^{\infty}\sum_{K = J}^{\infty} \frac{K}{X^K} =

∑ J = 1 ∞ ( J X J + J + 1 X J + 1 + J + 2 X J + 2 + . . . ) = \sum_{J = 1}^{\infty} (\frac{J}{X^J} + \frac{J + 1}{X^{J + 1}} + \frac{J + 2}{X^{J + 2}} + ...) =

∑ J = 1 ∞ J X J ∗ ( X X − 1 ) \sum_{J = 1}^{\infty} \frac{J}{X^J} * (\frac{X}{X - 1}) + + ∑ J = 1 ∞ 1 X J ∗ ( X ( X − 1 ) 2 ) ) \sum_{J = 1}^{\infty} \frac{1}{X^J} * (\frac{X}{(X - 1)^2})) = X 2 ( X − 1 ) 3 + X ( X − 1 ) 2 ∗ ( 1 X ) ∗ ( X X − 1 ) = = \frac{X^2}{(X - 1)^3} + \frac{X}{(X - 1)^2} * (\frac{1}{X}) * (\frac{X}{X - 1}) =

X 2 ( X − 1 ) 3 + X ( X − 1 ) 3 \frac{X^2}{(X - 1)^3} + \frac{X}{(X - 1)^3}

∑ N = 1 ∞ N 3 X N = \sum_{N = 1}^{\infty} \frac{N^3}{X^N} = ∑ J = 1 ∞ ∑ K = J ∞ K 2 X K = \sum_{J = 1}^{\infty}\sum_{K = J}^{\infty} \frac{K^2}{X^K} =

∑ J = 1 ∞ ( J 2 X J + J 2 + 2 J + 1 X J + 1 + J 2 + 4 J + 4 X J + 2 + . . . + J 2 + m ∗ J + m 2 X J + M + . . . ) \sum_{J = 1}^{\infty} (\frac{J^2}{X^J} + \frac{J^2 + 2J + 1}{X^{J + 1}} + \frac{J^2 + 4J + 4}{X^{J + 2}} + ... + \frac{J^2 + m * J + m^2}{X^{J + M}} + ...) = =

∑ J = 1 ∞ J 2 X J ∗ ( 1 + 1 X + 1 X 2 + . . . + 1 X m − 1 + . . . ) + \sum_{J = 1}^{\infty} \frac{J^2}{X^J} * (1 + \frac{1}{X} + \frac{1}{X^2} + ... + \frac{1}{X^{m - 1}} + ... ) + ∑ J = 1 ∞ 2 J X J ∗ ( 1 X + 2 X 2 + 3 X 3 + . . . + m X m + . . . ) \sum_{J = 1}^{\infty} \frac{2J}{X^J} * (\frac{1}{X} + \frac{2}{X^2} + \frac{3}{X^3} + ... + \frac{m}{X^m} + ...)

+ + ∑ J = 1 ∞ 1 X J ∗ ( 1 2 X + 2 2 X 2 + 3 2 X 3 + . . . + m 2 X m + . . . ) \sum_{J = 1}^{\infty} \frac{1}{X^J} * (\frac{1^2}{X} + \frac{2^2}{X^2} + \frac{3^2}{X^3} + ... + \frac{m^2}{X^m} + ...) = =

( X X − 1 ) ∗ ( ∑ J = 1 ∞ J 2 X J ) + ( 2 X ( X − 1 ) 2 ) ∗ ( ∑ J = 1 ∞ J X J ) + ( X 2 + X ( X − 1 ) 3 ) ∗ ( ∑ J = 1 ∞ 1 X J ) (\frac{X}{X - 1}) * (\sum_{J = 1}^{\infty} \frac{J^2}{X^J}) + (\frac{2X}{(X - 1)^2}) * (\sum_{J = 1}^{\infty} \frac{J}{X^J}) + (\frac{X^2 + X}{(X - 1)^3}) * (\sum_{J = 1}^{\infty} \frac{1}{X^J}) = =

( X X − 1 ) ∗ ( X 2 + X ( X − 1 ) 3 ) + ( 2 X ( X − 1 ) 2 ) ∗ ( X ( X − 1 ) 2 ) + ( X 2 + X ( X − 1 ) 3 ) ∗ ( 1 X ) ∗ ( X X − 1 ) = (\frac{X}{X - 1}) * (\frac{X^2 + X}{(X - 1)^3}) + (\frac{2X}{(X - 1)^2}) * (\frac{X}{(X - 1)^2}) + (\frac{X^2 + X}{(X - 1)^3}) * (\frac{1}{X}) * (\frac{X}{X - 1}) =

( X X − 1 ) ∗ ( X 2 + X ( X − 1 ) 3 ) + ( 2 X ( X − 1 ) 2 ) ∗ ( X ( X − 1 ) 2 ) + ( X 2 + X ( X − 1 ) 3 ) ∗ ( 1 X − 1 ) = (\frac{X}{X - 1}) * (\frac{X^2 + X}{(X - 1)^3}) + (\frac{2X}{(X - 1)^2}) * (\frac{X}{(X - 1)^2}) + (\frac{X^2 + X}{(X - 1)^3}) * (\frac{1}{X - 1}) = X 3 + 4 X 2 + X ( X − 1 ) 4 \frac{X^3 + 4X^2 + X}{(X - 1)^4} F o r ( X > 1 ) For (X > 1)

⟹ \implies

∫ 2 3 ∑ N = 1 ∞ N 3 x N d x = ∫ 2 3 X 3 + 4 X 2 + X ( X − 1 ) 4 d x \int_2^3 \sum_{N=1}^\infty \dfrac{N^3}{x^N} \, dx = \int_2^3 \frac{X^3 + 4X^2 + X}{(X - 1)^4} dx

L e t Let U = X − 1 U = X - 1 ⟹ \implies ∫ 1 2 U 3 + 7 U 2 + 12 U + 6 U 4 d u = ln ⁡ ( U ) − 7 U − 6 U 2 − 2 U 3 ∣ 1 2 \int_1^2 \frac{U^3 + 7U^2 + 12U + 6}{U^4} du = \ln(U) - \frac{7}{U} - \frac{6}{U^2} - \frac{2}{U^3}|_{1}^{2} = ln ⁡ ( 2 ) + 39 4 = \ln(2) + \frac{39}{4} = = 10.443147 10.443147 t o to s i x six d e c i m a l decimal p l a c e s . places.

e n d . end. A n o t h e r Another a p p r o a c h approach i s is t a k i n g taking d e r i v a t i v e s . derivatives.

F o r For e x a m p l e : example: T o To f i n d find ∑ N = 1 ∞ N 4 X N \sum_{N=1}^\infty \dfrac{N^4}{X^N} R e p l a c e Replace X X b y by 1 / X 1/X i n in ∑ 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} ⟹ \implies ∑ N = 1 ∞ ( N 3 ∗ X N ) = X 3 + 4 X 2 + X ( 1 − X ) 4 \sum_{N=1}^\infty (N^3 *X^N) = \frac{X^3 + 4X^2 + X}{(1 - X)^4}

d / d x ( ∑ N = 1 ∞ ( N 3 ∗ X N ) ) d/dx(\sum_{N=1}^\infty (N^3 * X^N)) = X 3 + 11 X 2 + 11 X + 1 ( 1 − X ) 5 = \frac{X^3 + 11X^2 + 11X + 1}{(1 - X)^5} ⟹ \implies ∑ N = 1 ∞ ( N 4 ∗ X N ) \sum_{N=1}^\infty (N^4 *X^N) = X ∗ ( X 3 + 11 X 2 + 11 X + 1 ) ( 1 − X ) 5 = \frac{X * (X^3 + 11X^2 + 11X + 1)}{(1 - X)^5} R e p l a c i n g Replacing X X b y by 1 / X 1/X w e we o b t a i n : 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}

2 Helpful 0 Interesting 1 Brilliant 0 Confused

The derivative approach to solving the series also makes the integral very easy to solve. You can just recognize that the series with N to a certain power in the numerator is equal to x times the derivative of the series with one lower power of N in the numerator and use that fact to repeatedly apply integration by parts and work backwards until you reach the final term.

Tristan Goodman - 1 year, 6 months ago

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Your right. On similar problems I had done I used derivatives. That's why I mentioned it at the end of my solution.

On a unrelated topic. I have been gone for about eight months. When I came back about two weeks ago I did twenty one new problems. I find it strange that I only received at most one response to each problem I posted. I.E; At most one person did each problem. It seems that no one is home. Is there a reason for this? I did notice Brilliant changed it's format.

Rocco Dalto - 1 year, 6 months ago

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I have never posted any problems of my own so I don't quite know what is considered normal with regards to the frequency of responses to new problems. Though as someone who frequents this site, I can't say that I've noticed any recent, significant change in activity. Was the daily problem section there before you were away? Perhaps it has diverted some attention away from the community problems.

Personally, I spend most of my time in the community section using the search bar to look up slightly more targeted types of problems that I happen to be interested in trying or practicing, rather than always looking at the new ones, so I don't always even see things as they are posted.

Tristan Goodman - 1 year, 6 months ago

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@Tristan Goodman – "Was the daily problem section there before you were away? "

No. It was a better format that included about 15 problems/week. It ranged from easy to moderate to difficult, where(from what I remember) each category contained five problems posted by users of brillant. There always was a lot of traffic. It's has changed. I myself posted 832 problems in the past couple of years. In any case, thanks for your response.

Rocco Dalto - 1 year, 6 months ago

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