Inverse Sum Squared

Calculus Level 5

∑ n = 1 ∞ ( ∑ k = 0 n k 2 ) − 1 \large\displaystyle\sum_{n=1}^{\infty}\left({\displaystyle\sum_{k=0}^nk^2}\right)^{-1} If the value of the series above can be expressed as a − b ln ⁡ ( c ) a-b\ln{(c)} where a , b a,b are positive integers and c c is the minimum possible positive integer, find the value of a + b + c a+b+c .

See Also Inverse Sum and Inverse Sum Cubed .


The answer is 44.

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Ikkyu San by Ikkyu San , 23, Thailand

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2 solutions

Chew-Seong Cheong
Sep 19, 2015

∑ n = 1 ∞ 1 ∑ k = 0 n k 2 = ∑ n = 1 ∞ 1 1 6 n ( n + 1 ) ( 2 n + 1 ) = ∑ n = 1 ∞ 6 n ( n + 1 ) ( 2 n + 1 ) = ∑ n = 1 ∞ 24 2 n ( 2 n + 2 ) ( 2 n + 1 ) = 12 ∑ n = 1 ∞ ( 1 2 n − 2 2 n + 1 + 1 2 n + 2 ) = 12 [ ( 1 2 − 2 3 + 1 4 ) + ( 1 4 − 2 5 + 1 6 ) + ( 1 6 − 2 7 + 1 8 ) + . . . ] = 12 [ 1 2 − 2 3 + 2 4 − 2 5 + 2 6 − 2 7 + 2 8 − . . . ] = 24 [ 1 2 − 1 3 + 1 4 − 1 5 + 1 6 − 1 7 + 1 8 − . . . ] − 6 = 24 − 24 [ 1 − 1 2 + 1 3 − 1 4 + 1 5 − 1 6 + 1 7 − 1 8 − . . . ] − 6 = 18 − 24 ln ⁡ 2 \begin{aligned} \sum_{n=1}^\infty \frac{1}{\sum_{k=0}^n k^2} & = \sum_{n=1}^\infty \frac{1}{\frac{1}{6}n(n+1)(2n+1)} \\ & = \sum_{n=1}^\infty \frac{6}{n(n+1)(2n+1)} \\ & = \sum_{n=1}^\infty \frac{24}{2n(2n+2)(2n+1)} \\ & = 12 \sum_{n=1}^\infty \left( \frac{1}{2n} - \frac{2}{2n+1} + \frac{1}{2n+2} \right) \\ & = 12 \left[ \left(\frac{1}{2} - \frac{2}{3} + \frac{1}{4} \right) + \left(\frac{1}{4} - \frac{2}{5} + \frac{1}{6} \right) + \left(\frac{1}{6} - \frac{2}{7} + \frac{1}{8} \right) + ... \right] \\ & = \color{#3D99F6}{12} \left[ \frac{\color{#D61F06}{1}}{2} - \frac{2}{3} + \frac{2}{4} - \frac{2}{5} + \frac{2}{6} - \frac{2}{7} + \frac{2}{8} - ... \right] \\ & = \color{#3D99F6}{24} \left[ \frac{\color{#D61F06}{1}}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \frac{1}{6} - \frac{1}{7} + \frac{1}{8} - ... \right] \color{#D61F06}{- 6} \\ & = \color{#3D99F6}{24} -24 \left[\color{#3D99F6}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} - ... \right] - 6 \\ & = 18 -24 \ln{2} \end{aligned}

⇒ a + b + c = 18 + 24 + 2 = 44 \Rightarrow a + b + c = 18+24+2 = \boxed{44}

11 Helpful 0 Interesting 1 Brilliant 0 Confused

Can you please correct me Chew-Seong Cheong sir

Akul Agrawal - 5 years, 8 months ago

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Akul Agrawa and Saurabh Chaturvedi , please note that ∑ n = 1 ∞ 1 r \sum_{n=1}^\infty \frac{1}{r} and ∑ n = 1 ∞ 1 r + 1 \sum_{n=1}^\infty \frac{1}{r+1} do not converge. We cannot split the terms into two summation and calculate. You see my solution only involves a summation.

Chew-Seong Cheong - 5 years, 8 months ago

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@Chew-Seong Cheong sir please tell me if rather than using summation directly in partial fraction...may I write 1/2n to integral of x^(2n-1) from 0 to 1 and all other subsequent terms like these and solve the geometric progression first then do the integral?

Righved K - 5 years, 6 months ago

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@Righved K – Sorry, I am not quite sure you are asking. Please learn up LaTex. You can just put "\" and "(" (with no space between them) before and "\" and ")" and after the formulas. You can place your mouse cursor on top of formulas in the question and solution parts of this site to find out the keystrokes.

Chew-Seong Cheong - 5 years, 6 months ago
Akul Agrawal
Sep 19, 2015

answer is ∑ r = 1 ∞ 6 ∗ [ 1 r − 4 2 r + 1 + 1 r + 1 ] \sum _{ r=1 }^{ \infty }{ 6*\left[ \frac { 1 }{ r } -\frac { 4 }{ 2r+1 } +\frac { 1 }{ r+1 } \right] }

=> 6 [1+2 ( (1/2)+(1/3)+(1/4) ...)-4*( (1/3)+(1/5)+(1/7) ...)]

=> 6 [1+ 2 ( (1/2)-(1/3)+(1/4)-(1/5) ...)]

=>6 [1+2 ( -log(2)+1)]

=>6*[3-2log(2)]=18-12log(2)

1 Helpful 1 Interesting 0 Brilliant 0 Confused

Even I was getting the same answer. 32. What's wrong?

Saurabh Chaturvedi - 5 years, 8 months ago

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