Change in change in harmony

Calculus Level 4

∑ n = 1 ∞ [ H n ( 2 ) ( 1 n 2 − 1 ( n + 1 ) 2 ) ] \large \sum_{n=1}^\infty \left [ H_n^{(2)} \left ( \dfrac1{n^2} - \dfrac1{(n+1)^2} \right) \right ]

If the value of the series above is in the form of π A B \dfrac{\pi^ A}B , where A A and B B are integers, find A + B A+B .

Notation : H n ( m ) H_n^{(m)} denotes the generalized harmonic number, H n ( m ) = ∑ k = 1 n 1 k m \displaystyle H_n^{(m)} = \sum_{k=1}^n \dfrac1{k^m} .

Inspirations: First link , Second link .

There is a much simpler way to solve this than the two other questions.


The answer is 94.

Joel Yip by Joel Yip , 21, Singapore

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Joel Yip
Mar 11, 2016

First

∑ n = 1 ∞ H n ( 2 ) ( 1 n 2 − 1 ( n + 1 ) 2 ) = ∑ n = 1 ∞ H n ( 2 ) n 2 − H n ( 2 ) ( n + 1 ) 2 { \sum _{ n=1 }^{ \infty }{ { { H }_{ n }^{ (2) } }\left( \frac { 1 }{ { n }^{ 2 } } -\frac { 1 }{ { \left( n+1 \right) }^{ 2 } } \right) } }=\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } -\frac { { H }_{ n }^{ (2) } }{ { \left( n+1 \right) }^{ 2 } } }

then we will try to make it a telescoping series.

∑ n = 1 ∞ H n ( 2 ) n 2 − H n ( 2 ) ( n + 1 ) 2 = ∑ n = 1 ∞ H n ( 2 ) n 2 − H n ( 2 ) + 1 ( n + 1 ) 2 ( n + 1 ) 2 + 1 ( n + 1 ) 2 ( n + 1 ) 2 ] \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } -\frac { { H }_{ n }^{ (2) } }{ { \left( n+1 \right) }^{ 2 } } } =\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } -\frac { { H }_{ n }^{ (2) }+\frac { 1 }{ { \left( n+1 \right) }^{ 2 } } }{ { \left( n+1 \right) }^{ 2 } } +\frac { \frac { 1 }{ { \left( n+1 \right) }^{ 2 } } }{ { \left( n+1 \right) }^{ 2 } } } ]

By definition

H n ( 2 ) + 1 ( n + 1 ) 2 = H n + 1 ( 2 ) { H }_{ n }^{ \left( 2 \right) }+\frac { 1 }{ { \left( n+1 \right) }^{ 2 } } ={ H }_{ n+1 }^{ \left( 2 \right) }

so,

∑ n = 1 ∞ H n ( 2 ) n 2 − H n + 1 ( 2 ) ( n + 1 ) 2 + 1 ( n + 1 ) 2 ( n + 1 ) 2 \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } -\frac { { H }_{ n+1 }^{ \left( 2 \right) } }{ { \left( n+1 \right) }^{ 2 } } +\frac { \frac { 1 }{ { \left( n+1 \right) }^{ 2 } } }{ { \left( n+1 \right) }^{ 2 } } }

∑ n = 1 ∞ H n ( 2 ) n 2 − H n + 1 ( 2 ) ( n + 1 ) 2 + 1 ( n + 1 ) 2 ( n + 1 ) 2 = ∑ n = 1 ∞ H n ( 2 ) n 2 − ∑ n = 1 ∞ H n + 1 ( 2 ) ( n + 1 ) 2 + ∑ n = 1 ∞ 1 ( n + 1 ) 2 ( n + 1 ) 2 \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } -\frac { { H }_{ n+1 }^{ \left( 2 \right) } }{ { \left( n+1 \right) }^{ 2 } } +\frac { \frac { 1 }{ { \left( n+1 \right) }^{ 2 } } }{ { \left( n+1 \right) }^{ 2 } } } =\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } } -\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n+1 }^{ \left( 2 \right) } }{ { \left( n+1 \right) }^{ 2 } } } +\sum _{ n=1 }^{ \infty }{ \frac { \frac { 1 }{ { \left( n+1 \right) }^{ 2 } } }{ { \left( n+1 \right) }^{ 2 } } }

∑ n = 1 ∞ H n ( 2 ) n 2 − ∑ n = 1 ∞ H n + 1 ( 2 ) ( n + 1 ) 2 + ∑ n = 1 ∞ 1 ( n + 1 ) 2 ( n + 1 ) 2 = ∑ n = 1 ∞ H n ( 2 ) n 2 − ∑ n = 1 ∞ H n + 1 ( 2 ) ( n + 1 ) 2 + ∑ n = 1 ∞ 1 ( n + 1 ) 4 \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } } -\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n+1 }^{ \left( 2 \right) } }{ { \left( n+1 \right) }^{ 2 } } } +\sum _{ n=1 }^{ \infty }{ \frac { \frac { 1 }{ { \left( n+1 \right) }^{ 2 } } }{ { \left( n+1 \right) }^{ 2 } } } =\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } } -\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n+1 }^{ \left( 2 \right) } }{ { \left( n+1 \right) }^{ 2 } } } +\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { \left( n+1 \right) }^{ 4 } } }

∑ n = 1 ∞ H n ( 2 ) n 2 − ∑ n = 1 ∞ H n + 1 ( 2 ) ( n + 1 ) 2 \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } } -\sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n+1 }^{ \left( 2 \right) } }{ { \left( n+1 \right) }^{ 2 } } } is a telescoping series,

∑ n = 1 ∞ H n ( 2 ) n 2 − ∑ n = 2 ∞ H n ( 2 ) n 2 \sum _{ n=1 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } } -\sum _{ n=2 }^{ \infty }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } }

so ∑ n = 1 1 H n ( 2 ) n 2 = 1 \sum _{ n=1 }^{ 1 }{ \frac { { H }_{ n }^{ (2) } }{ { n }^{ 2 } } =1 }

1 + ∑ n = 1 ∞ 1 ( n + 1 ) 4 = 1 + ∑ n = 1 ∞ 1 n 4 − 1 1+\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { \left( n+1 \right) }^{ 4 } } } =1+\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 4 } } } -1

1 + ∑ n = 1 ∞ 1 n 4 − 1 = 1 + ζ ( 4 ) − 1 1+\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 4 } } } -1=1+\zeta \left( 4 \right) -1

1 + ζ ( 4 ) − 1 = ζ ( 4 ) = π 4 90 1+\zeta \left( 4 \right) -1=\zeta \left( 4 \right) =\frac { { \pi }^{ 4 } }{ 90 }

so 4 + 90 = 94 4+90=\boxed{94}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice way of using both the summations together!

Aditya Kumar - 5 years, 3 months ago

Log in to reply

thank you for that

Joel Yip - 5 years, 3 months ago

S = ∑ n = 1 ∞ [ H n ( 2 ) ( 1 n 2 − 1 ( n + 1 ) 2 ) ] = ∑ n = 1 ∞ H n ( 2 ) n 2 − ∑ n = 1 ∞ H n ( 2 ) ( n + 1 ) 2 Check Inspiration 1 and Inspiration 2 = ∑ n ≥ m ≥ 1 1 m 2 n 2 − ∑ n > m ≥ 1 1 m 2 n 2 = ∑ n = m 1 m 2 n 2 = ∑ n = 1 ∞ 1 n 4 = ζ ( 4 ) = π 4 90 \begin{aligned} S & = \sum_{n=1}^\infty \left[ H_n^{(2)}\left(\frac{1}{n^2} - \frac{1}{(n+1)^2} \right) \right] \\ & = \color{#3D99F6}{ \sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2}} - \color{#D61F06}{ \sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2}} \quad \quad \small \text{Check } \color{#3D99F6}{\text{Inspiration 1 }} \text{and } \color{#D61F06}{\text{Inspiration 2 }} \\ & = \color{#3D99F6}{ \sum_{n \ge m \ge 1} \frac{1}{m^2n^2}} - \color{#D61F06}{ \sum_{n > m \ge 1} \frac{1}{m^2n^2}} \\ & = \sum_{n = m} \frac{1}{m^2n^2} = \sum_{n=1}^\infty \frac{1}{n^4} = \zeta (4) = \frac{\pi^4}{90} \end{aligned}

⟹ A + B = 4 + 90 = 94 \implies A+B = 4+90 = \boxed{94}

Inspirations for the solution are from:

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...