Hey guys was wondering if someone could help me figure out what is happening in this practice problem as I cant determine what they are doing certain steps. $$\sum_{n=1}^\infty \frac{6}{9n^2+3n-2}$$ The top is the original problem, after factoring the denominator you get the following $$\sum_{n=1}^{\infty}\frac{6}{(3n-1)(3n+2)}$$ after this the example does the following $$\sum_{n=1}^\infty \left[\frac{6}{9n-3}-\frac{6}{9n+6}\right]=2\sum_{n=1}^ \infty \left[\frac{1}{3n-1}-\frac{1}{3n+2}\right]$$ Now my question is how they went from increasing the denominator by a factor of 3 and removing the 2. I dont understand why and how that was done if someone could please explain this I would greatly appreciate it.
2026-03-29 12:05:31.1774785931
Factoring out a value to find the convergent series.
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This comes from partial fractions decomposition: $$\frac1{(3n-1)(3n+2)}=\frac13\frac 1{3n-1}-\frac13\frac 1{3n+2}.$$ I don't think it was a good idea to expand the denominators.
Some details:
We can decompose $$\frac1{(3x-1)(3x+2)}=\frac A{3x-1}+\frac B{3x+2}$$ for a unique pair of numbers $(A,B)$. Multiply both sides by $(3x-1)(3x+2)$ to remove denominators. We obtain $$1=A(3x+2)+B(3x-1)$$ Now set $x=\dfrac13$. This yields $\;1=A\cdot(1+2)+B\cdot 0$, whence $\;A=\dfrac13$.
Similarly, set $x=-\dfrac23$. You get $\;1=A\cdot 0+B(-2-1)$, whence $B=-\dfrac13$.