Time for easier problems

Calculus Level 4

∑ k = 1 ∞ sin ⁡ k k 2 = A B i ( Li 2 ( e − C i ) − Li 2 ( e D i ) ) \large \displaystyle\sum _{ k=1 }^{ \infty }{ \frac { \sin { k } }{ { k }^{ 2 } } =\frac { A }{ B } i({ \text{Li} }_{ 2 }({ e }^{ -Ci } }) -{\text{Li} }_{ 2 }({ e }^{ Di }))

The equation above holds true for positive integers A , B , C A,B,C and D D , with A , B A,B coprime. Find A + B + C + D A+B+C+D .

Notation : Li n ( a ) { \text{Li} }_{ n }(a) denotes the polylogarithm function, Li n ( a ) = ∑ k = 1 ∞ a k k n . { \text{Li} }_{ n }(a)=\displaystyle\sum _{ k=1 }^{ \infty }{ \frac { { a }^{ k } }{ { k }^{ n } } }.

Clarification : i = − 1 i=\sqrt{-1}


The answer is 5.

Hamza A by Hamza A , 19, USA

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1 solution

Aditya Kumar
Mar 8, 2016

We use: sin ⁡ ( k ) = e i k − e − i k 2 i \sin { \left( k \right) } =\frac { { e }^{ ik }-{ e }^{ -ik } }{ 2i }

∑ k = 1 ∞ sin ⁡ ( k ) k 2 = ∑ k = 1 ∞ e i k − e − i k 2 i k 2 \sum _{ k=1 }^{ \infty }{ \frac { \sin { \left( k \right) } }{ { k }^{ 2 } } } =\sum _{ k=1 }^{ \infty }{ \frac { { e }^{ ik }-{ e }^{ -ik } }{ 2i{ k }^{ 2 } } }

∑ k = 1 ∞ sin ⁡ ( k ) k 2 = i 2 ∑ k = 1 ∞ − e i k + e − i k k 2 \sum _{ k=1 }^{ \infty }{ \frac { \sin { \left( k \right) } }{ { k }^{ 2 } } } =\frac { i }{ 2 } \sum _{ k=1 }^{ \infty }{ \frac { -{ e }^{ ik }+{ e }^{ -ik } }{ { k }^{ 2 } } }

∑ k = 1 ∞ sin ⁡ ( k ) k 2 = i 2 { L i 2 ( e − i ) − L i 2 ( e i ) } \sum _{ k=1 }^{ \infty }{ \frac { \sin { \left( k \right) } }{ { k }^{ 2 } } } =\frac { i }{ 2 } \left\{ { Li }_{ 2 }\left( { e }^{ -i } \right) -{ Li }_{ 2 }\left( { e }^{ i } \right) \right\}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

exact intended solution,upvoted! :)

Hamza A - 5 years, 3 months ago

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Level 5 makes me laugh :D

Aditya Kumar - 5 years, 3 months ago

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why is this showing as level 5?shouldn't it change if i made the wrong level for the problem,if it stays like that,learn some Euler guys!

Hamza A - 5 years, 3 months ago

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