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When I put this into wolfram alpha it will not calculate it because it takes too long, maybe someone who has wolfram alpha pro can help me or maybe someone can solve that just by his know-how.

Let f be a number lets says 100 kHz so 100000 for example or 150000 (does this change anyway the result in the end if its 100k or 150k ?)

The series is the following:

$$\sum_{k=0}^\infty \frac{2\cdot(-1)^{k+1}}{\pi(4k^2-1)}\cdot \cos(4\pi kft)$$

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There is a solution in terms of hypergeometric functions since (I removed the $\pi$ factor) $$\sum_{k=0}^\infty \frac{2\,(-1)^{k+1}}{4k^2-1}\, \cos(4\pi kft)=\, _2F_1\left(-\frac{1}{2},1;\frac{3}{2};-e^{-4 i \pi f t}\right)+\, _2F_1\left(-\frac{1}{2},1;\frac{3}{2};-e^{4 i \pi f t}\right)$$

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Why do you think this would be useful?

$$ \sum _{k=0}^{\infty }2\,{\frac { \left( -1 \right) ^{k+1}\cos \left( 4 \,\pi \,kft \right) }{\pi \, \left( 4\,{k}^{2}-1 \right) }} \\={ \frac {-1}{4 \pi \, \left( {{\rm e}^{4\,i\pi \,ft}}+1 \right) } \left[ i{ {\rm e}^{6\,i\pi \,ft}}\ln \left( {\frac {-{{\rm e}^{2\,i\pi \,ft}}+i }{{{\rm e}^{2\,i\pi \,ft}}+i}} \right) +2\,i{{\rm e}^{2\,i\pi \,ft}} \ln \left( {\frac {-{{\rm e}^{2\,i\pi \,ft}}+i}{{{\rm e}^{2\,i\pi \,f t}}+i}} \right)\\ -4\,{{\rm e}^{4\,i\pi \,ft}}+i{{\rm e}^{6\,i\pi \,ft}} \ln \left( -{\frac {{{\rm e}^{2\,i\pi \,ft}}+i}{-{{\rm e}^{2\,i\pi \, ft}}+i}} \right) +2\,i{{\rm e}^{2\,i\pi \,ft}}\ln \left( -{\frac {{ {\rm e}^{2\,i\pi \,ft}}+i}{-{{\rm e}^{2\,i\pi \,ft}}+i}} \right)\\ +i{ {\rm e}^{-2\,i\pi \,ft}}\ln \left( {\frac {-{{\rm e}^{2\,i\pi \,ft}}+ i}{{{\rm e}^{2\,i\pi \,ft}}+i}} \right) +i{{\rm e}^{-2\,i\pi \,ft}} \ln \left( -{\frac {{{\rm e}^{2\,i\pi \,ft}}+i}{-{{\rm e}^{2\,i\pi \, ft}}+i}} \right) -4 \right] } $$ from Maple.

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