<s>[INST] Sum of an infinite series $\sum _{k=3}^{\infty }\:\frac{1}{k\left(k^4-5k^2+4\right)^2}$ I want to compute this series, $$\sum _{k=3}^{\infty }\:\frac{1}{k\left(k^4-5k^2+4\right)^2}.$$ I don't know how to contiune after factoring, which yields $\frac{1}{k(k-1)^2(k+1)^2(k-2)^2(k+2)^2}.$
I know that $\frac{1}{k(k-1)}$ telescopes. However, I can't apply this here, as the product of the sum is not equal to the sum of the product. I only want hints for now.
 [/INST] Well, you seem to know how to telescope summing:
$$\frac1{k(k-1)} = \frac1{k-1} - \frac1k$$
Similarly, note
$$\frac1{(k-2)^2(k+2)^2} = \frac1{8k}\left(\frac1{(k-2)^2}- \frac1{(k+2)^2}\right)$$
$$\implies \frac1{(k-2)^2(k-1)^2k(k+1)^2(k+2)^2} = \frac1{8(k-1)^2k^2(k+1)^2}\left(\frac1{(k-2)^2}- \frac1{(k+2)^2}\right)$$
which can be noted to telescope, hence the sum is $\dfrac1{8\cdot 2^2\cdot3^2\cdot4^2}\cdot\dfrac1{1^2}=\dfrac1{4608}$.
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