Do I need to use partial fractions to find $\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$?

Question

I need to simplify $\frac{2n+1}{n^2(n+1^2)}$ as part of an exam question. The solution states $$\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$$ In the solution it does not state how this simplification was made, I figured this could be done in quite a long winded fashion, using partial fractions. But from how it's written in the solutions it seems like this should be an easy simplification.

Is there a simple trick to simplifying fractions like this?

Answer 1

Note that $$\frac{2n+1}{n^2(n+1)^2} = \frac{(n^2 + 2n + 1)-n^2}{n^2(n+1)^2}$$ $$=\frac{(n+1)^2-n^2}{n^2(n+1)^2}$$ $$=\frac{(n+1)^2}{n^2(n+1)^2}-\frac{n^2}{n^2(n+1)^2}$$ $$=\frac1{n^2}-\frac1{(n+1)^2}$$ as desired. This approach is motivated by noticing that the original numerator is "almost" a recognizable perfect square, so one is tempted to add and subtract the missing quantity that completes the square.

Answer 2

HINT

Use Partial Fraction Decomposition,

$$\frac{2n+1}{n^2(n+1)^2}=\frac An+\frac B{n^2}+\frac C{n+1}+\frac D{(n+1)^2}$$

$$\iff2n+1=An(n+1)^2+B(n+1)^2+Cn^2(n+1)+Dn^2$$

Now compare the constants & the coefficients of $n,n^2,n^3$ to find $A,B,C,D$

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Alternatively by observation,

$$\frac{2n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{(n+1)^2n^2}=\cdots$$

Answer 3

use that $\frac{2n+1}{n^2(n+1)^2}=\frac{A}{n}+\frac{B}{n^2}+\frac{C}{n+1}+\frac{D}{(n+1)^2}$