Much ado about nothing

Let $\dfrac{k - 3}{k(k^{2} - 1)} = \dfrac{A}{k} + \dfrac{B}{k - 1} + \dfrac{C}{k + 1}.$

Then $k - 3 = A(k^{2} - 1) + Bk(k + 1) + Ck(k - 1) = (A + B + C)k^{2} + (B - C)k - A.$

Comparing like coefficients, we see that $A + B + C = 0, B - C = 1$ and $A = 3.$

Thus $(B + C) + (B - C) = -3 + 1 = -2 \Longrightarrow B = -1$ and $C = -2.$

So $\dfrac{k - 3}{k(k^{2} - 1)} = \dfrac{3}{k} - \dfrac{1}{k - 1} - \dfrac{2}{k + 1}.$

Now since the series clearly converges by the comparison test with $\sum \frac{1}{k^{2}}$ we are free to rearrange terms. In doing so, we note that each integer denominator $n \ge 3$ will appear (in unreduced fractions) three times, once for each of the fractions in the above sum. In each case this rearrangement will sum to $0,$ leaving the occurrences of $1$ and $2$ in denominators as the only fractions we need to consider further. Now we have one occurrence of denominator $1$ when $k = 2$ and two occurrences of denominator $2,$ one each for $k = 2$ and $k = 3.$ These sum to

$\dfrac{3}{2} - \dfrac{1}{2 - 1} - \dfrac{1}{3 - 1} = \boxed{0}.$

Much ado about nothing, indeed. :) (The title was a pretty strong hint.)

Express the given fraction in terms of partial fractions & apply cover-up rule to get the expression (k-3)/(k^2-1)=-1/(k-1) + 3/k - 2/(k+1).

From there apply the method of difference on the summation where infinity is replaced by N to obtain 1/N - 2/(N+1).

From there use the concept of reciprocal of N tending to zero as N tends to infinity to obtain zero.

(k-3)/ k(k^2 -1) = 1/2 ( 1/k-1 - 1/k+1 ) - 1/2 { 3/(k-1)k - 3/k(k+1) } So Ans) = 1/2 ( 1/1 + 1/2 ) - 1/2 ( 3/1*2 ) = 3/4 - 3/4 = 0 (by telescoping sum )