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$$\sum_{n=1}^\infty \frac{n}{4n^4+1}$$ my attempt : assumed the series is a telescopic and tried finding $t_n - t_{n-1}$ but then realized it is not a telescopic series. $$$$ //answer is given to be 0.25

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closed as off-topic by Saad, Javi, José Carlos Santos, Paul Frost, Wouter Apr 18 at 14:02

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Use the factorization: $$\dfrac{n}{4n^4+1} = \dfrac{n}{(2n^2-2n+1)(2n^2+2n+1)} = \frac 14\cdot\dfrac{1}{2n^2-2n+1}-\frac 14\cdot\dfrac{1}{2n^2+2n+1}$$

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