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Find $\sum_1^n$ $\frac {2r+1}{r^2(r+1)^2}$ Also, find the sum to infinity of the series. I tried decomposing it into partial fractions of the form $\frac Ar$ + $\frac{B}{r^2}$ + $\frac{C}{(r+1)}$ + $\frac{D}{(r+1)^2}$ but it was getting too complicated and tedious. Is there some trick here that i'm missing?

user140161
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2 Answers2

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$$ \frac1{r^2} - \frac1{(r+1)^2} = \frac{(r+1)^2-r^2}{r^2(r+1)^2} = \frac{(r^2+2r+1)-r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2} $$

The thing that suggested this to me is that $\displaystyle 2r+1 = 2\left(r+\frac12\right)$, and $r+\dfrac12$ is half-way between $r+0$ and $r+1$, the two expressions in the denominator.

Next, you rely on the fact that the sum telescopes, so nearly all of the terms vanish.

Michael Hardy
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You're on the right track! Clearing the fractions, we have: $$ 2r + 1 = Ar(r + 1)^2 + B(r + 1)^2 + Cr^2(r + 1) + Dr^2 $$ Comparing coefficients, we have: \begin{align*} \boxed{r^3}:\qquad 0 &= A + C \\ \boxed{r^2}:\qquad 0 &= 2A + B + C + D \\ \boxed{r^1}:\qquad 2 &= A + 2B \\ \boxed{r^0}:\qquad 1 &= B \\ \end{align*} Substituting, notice that $B = 1 \implies A = 0 \implies C = 0 \implies D = -1$. Hence, we obtain a telescoping series: $$ \sum_{r=1}^\infty \frac {2r+1}{r^2(r+1)^2} = \sum_{r=1}^\infty \left[\frac{1}{r^2} - \frac{1}{(r + 1)^2} \right] = \lim_{n\to\infty} \left[\frac{1}{1^2} - \frac{1}{(n + 1)^2} \right] = 1 $$

Adriano
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