Evaluate $\sum_{r=1}^{\infty} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$

Question

I was only able to observe that:

$\dfrac{r^2 - 1}{r^4 + r^2 + 1} = \dfrac{r^2 - 1}{(r^2 + r + 1)(r^2 - r + 1)}$

This hints at telescoping, but I would need an $r$ term in the numerator.

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The original question was

Evaluate $\sum_{r=1}^{\infty} \dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)}$

I was able to simplify it to the following:

$\dfrac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)} = \dfrac{(r^4 + r^3 - r^2 - r)}{(r^4 + r^2 + 1)(r^2 + r)} + \dfrac{3r^2+r+1}{\{(r+1)(r^2 + r + 1)\}\{r(r^2 - r + 1)\}} = \dfrac{r^2 - 1}{r^4 + r^2 + 1} + \left[\dfrac{1}{r(r^2 - r + 1)} - \dfrac{1}{(r+1)(r^2 + r + 1)}\right]$

The second term can be evaluated using telescoping, and the first term is what this post is asking for.

Any other ways of solving the original question are also welcome.

Answer 1

As this Art of Problem Solving thread suggests, write this using a form of partial fractions:

$$(Ar + B)(r^2 - r + 1) + (Cr + D)(r^2 + r + 1) = r^2 - 1$$

Now comparing the coefficients of $r^3$, $r^2$, $r$, and the constant term, we end up with $A + C = 0, B - A + D + C = 1, -B + A+D+C=0$, and $B + D = -1$.

Adding the middle two equations gives $D + C = 1/2$, and so $B - A = 1/2$. Now $B + (C) = 1/2$ and so we obtain $2(B + C + D) = 0$, which leads to $B = -1/2$, $C = 1$, $D = -1/2$, and $A = -1$.

Hence the summand can be written as:

$$\frac{-r - 1/2}{r^2 + r + 1} + \frac{r-1/2}{r^2 - r + 1}$$

which is telescoping.

Answer 2

If you are comfortable with generalized harmonic number, you could consider first the partial sum $$S_n=\sum_{r=1}^{n} \dfrac{r^2 - 1}{r^4 + r^2 + 1}$$ and write first $$\dfrac{r^2 - 1}{r^4 + r^2 + 1}=\frac{(r-1)(r+1)}{(r-a)(r-b)(r-c)(r-d)}$$ Using partial fraction decomposition, this is $$\frac{a^2-1}{(a-b) (a-c) (a-d) (r-a)}+\frac{b^2-1}{(b-a) (b-c) (b-d) (r-b)}+$$ $$\frac{c^2-1}{(c-a) (c-b) (c-d) (r-c)}+\frac{d^2-1}{(d-a) (d-b) (d-c) (r-d)}$$ and use $$\sum_{r=1}^n \frac 1{r-k}=H_{n-k}-H_{-k}$$

After simplification $$S_n=\frac{(n-1) n}{2 \left(n^2+n+1\right)}$$

Answer 3

Write \begin{align}&\frac{r^2-1}{r^4+r^2+1} = \frac{Ar+B}{r^2-r+1}+\frac{Cr+D}{r^2+r+1}\\ \iff &r^2-1 = (Ar+B)(r^2+r+1)+(Cr+D)(r^2-r+1)\tag{1}\end{align}

You can now expand the RHS as a polynomial in $r$ and by direct comparison of coefficients on the LHS and the RHS get a $4\times4$ linear system.

Alternatively, let $\alpha$ be a root of $x^2-x+1$. In particular, we have $\alpha^2 = \alpha - 1$. Evaluating $(1)$ at $\alpha$ gives us

\begin{align}\alpha^2-1 &= (A\alpha + B)(\alpha^2+\alpha+1)\\ (\alpha-1)-1 &= (A\alpha + B)((\alpha-1)+\alpha+1)\\ \alpha - 2 &= 2\alpha(A\alpha+B)\\ \alpha - 2 &= 2A\alpha^2+2B\alpha\\ \alpha - 2 &= 2A(\alpha - 1)+2B\alpha\\ \alpha - 2 &= 2(A+B)\alpha-2A \implies 2(A+B)=1,\ -2A = -2 \end{align}

Note that it's important to get rid of any higher powers of $\alpha$ besides $\alpha$ and $1$ before you compare coefficients. Similarly, if you take $\beta$ a root of $x^2+x+1$, you can calculate $C$ and $D$ analogously.

Now that you can calculate partial fractions decomposition, we have

\begin{align} \sum_{r = 1}^n \frac{r^2-1}{r^4+r^2+1} &=\sum_{r = 1}^n\left( \frac{2r-1}{2(r^2-r+1)}-\frac{2r+1}{2(r^2+r+1)} \right) \\ &= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 1}^n \frac{2r+1}{2(r^2+r+1)}\\ &= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 1}^n \frac{2(r+1)-1}{2((r+1)^2-(r+1)+1)} \\ &= \sum_{r = 1}^n \frac{2r-1}{2(r^2-r+1)} - \sum_{r = 2}^{n+1} \frac{2r-1}{2(r^2-r+1)} \\ &= \frac 12 - \frac{2n+1}{2(n^2+n+1)} \end{align}

Taking a limit as $n\to \infty$ gives us $$\sum_{r = 1}^\infty \frac{r^2-1}{r^4+r^2+1} = \frac 12.$$

Answer 4

$$r^4+r^2+1=(r^2+1)^2-r^2=(r^2+r+1)(r^2-r+1)$$

Observe that if $f(n)=n^2+n+1, f(r-1)=(r-1)^2-(r-1)+1=r^2-r+1$

If $g(n)=an+b, g(n-1)=a(n-1)+b=an+b-a$

$$\text{Let }\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(r)}{f(r)}-\dfrac{g(r-1)}{f(r-1)}$$ $$\text{so that the partial sum becomes }\sum_{r=1}^n\dfrac{r^2-1}{r^4+r^2+1}=\dfrac{g(n)}{f(n)}-\dfrac{g(0)}{f(0)}$$

$$\text{Now, }\lim_{n\to\infty}\dfrac{g(n)}{f(n)}=\lim_{n\to\infty}\dfrac{an+b}{n^2+n+1}=\lim_{n\to\infty}\dfrac{\dfrac an+\dfrac b{n^2}}{1+\dfrac1n+\dfrac1{n^2}}=0\text{ for finite a,b}$$

$$\text{ and } \dfrac{g(0)}{f(0)}=?\text{ so we don't need the actual value of }a$$

$$\begin{align}\implies r^2-1=(ar+b)(r^2-r+1)-(ar+b-a)(r^2+r+1)=-ar^2+(a-2b)r+a\end{align}$$

$$\implies -a=1\iff a=-1, a-2b=0\iff b=\dfrac a2=-\dfrac12$$