Find the sum of the series $\sum \limits_{n=3}^{\infty} \dfrac{1}{n^5-5n^3+4n}$ Feel free to skip obvious steps, or use a calculator when required.
I just want to understand the theme of the solution.
Any help is appreciated

EDIT :
We can write$$ \dfrac{1}{n^5-5n^3+4n} =  -\dfrac{1}{6 (n-1)}+\dfrac{1}{4 n}-\dfrac{1}{ 6(n+1)}+\dfrac{1}{ 24(n+2)}+\dfrac{1}{ 24(n-2)}$$
How do I do telescopic sum !
 A: $$\sum_{n=3}^{\infty}\dfrac{1}{n^5-5n^3+4n} = \sum_{n=3}^{\infty}\dfrac{1}{(n-2)(n-1)n(n+1)(n+2)}$$
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Rewrite this as $$  \dfrac{1}{4}\sum_{n=3}^{\infty}\dfrac{(n+2)-(n-2)}{(n-2)(n-1)n(n+1)(n+2)}$$
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$$  \dfrac{1}{4}\sum_{n=3}^{\infty}\left[\dfrac{(n+2)}{(n-2)(n-1)n(n+1)(n+2)}-\dfrac{(n-2)}{(n-2)(n-1)n(n+1)(n+2)} \right]$$
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$$  \dfrac{1}{4}\sum_{n=3}^{\infty}\left[\dfrac{1}{(n-2)(n-1)n(n+1)}-\dfrac{1}{(n-1)n(n+1)(n+2)} \right]$$
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$$\textbf{This is a telescoping sum}$$
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$$\lim\limits_{n \to \infty}\dfrac{1}{4} \sum_{i=3}^{n}\left[\dfrac{1}{(i-2)(i-1)i(i+1)}-\dfrac{1}{(i-1)i(i+1)(i+2)} \right]$$
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$$\require{cancel} =\lim\limits_{n \to \infty} \dfrac{1}{4}\left[\dfrac{1}{1\cdot 2\cdot 3\cdot 4} \cancel{-\dfrac{1}{2\cdot 3\cdot 4\cdot 5}  } \right]$$
$$ + \dfrac{1}{4}\left[\cancel{\dfrac{1}{2\cdot 3\cdot 4\cdot 5}}\cancel{ -\dfrac{1}{3\cdot 4\cdot 5\cdot 6}  } \right]$$
$$ + \dfrac{1}{4}\left[\cancel{\dfrac{1}{3\cdot 4\cdot 5\cdot 6}}\cancel{ -\dfrac{1}{4\cdot 5\cdot 6\cdot 7}   }\right]$$
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........
$$ + \dfrac{1}{4}\left[\cancel{\dfrac{1}{(n-4)\cdot (n-3)\cdot (n-2)(n-1)} }\cancel{-\dfrac{1}{(n-3)\cdot (n-2)\cdot (n-1)\cdot(n)} }  \right]$$
$$ + \dfrac{1}{4}\left[\cancel{\dfrac{1}{(n-3)\cdot (n-2)\cdot (n-1)\cdot (n)}} -\dfrac{1}{(n-2)\cdot (n-1)\cdot n\cdot (n+1)}   \right]$$
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$$ =\lim\limits_{n \to \infty} \dfrac{1}{4}\left[\dfrac{1}{1\cdot 2\cdot 3\cdot 4} -\dfrac{1}{(n-2)\cdot (n-1)\cdot n\cdot (n+1)}   \right]$$
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$$ = \dfrac{1}{4}\left[\dfrac{1}{1\cdot 2\cdot 3\cdot 4} -0  \right] = \dfrac{1}{1\cdot 2\cdot 3\cdot 4^2} = \dfrac{1}{96}$$
