Question

$\sum_{r=1}^{\infty} \frac{r^{3}+\left(r^{2}+1\right)^{2}}{\left(r^{4}+r^{2}+1\right)\left(r^{2}+r\right)}$ is equal to (A) \(3 / 2\) (B) 1 (C) 2 (D) infinite

Short answer

Answer: The values of the coefficients are \(A = 1\), \(B = -1\), and \(C = -1\).

Step-by-step solution

Identify the rational function and the partial fraction decomposition

Let's call the fraction inside the summation \(f(r)\). The given series can be rewritten as \(\sum_{r=1}^{\infty} f(r)\), where $$ f(r) = \frac{r^3 + (r^2 + 1)^2}{(r^4 + r^2 + 1)(r^2 + r)}. $$ Our objective is to rewrite \(f(r)\) as a sum of simpler fractions. We can do this using the partial fraction decomposition method. Specifically, we want to find constants \(A, B, C\) such that $$ f(r) = \frac{A}{r} + \frac{B}{r + 1} + \frac{C}{r^4 + r^2 + 1}. $$

Find the partial fraction decomposition

Clear the denominator by multiplying both sides by the least common multiple of the denominators, which is \((r)(r+1)(r^4+r^2+1)\), to get: $$ r^3 + (r^2 + 1)^2 = A(r + 1)(r^4 + r^2 + 1) + B(r)(r^4 + r^2 + 1) + C(r)(r + 1). $$ Now, we need to find the values of \(A, B, C\) so that this equation holds true for every value of \(r\). We do this by plugging in values of \(r\). Start with \(r = 0\), which gives \(A = 1\). Next, plug in \(r = -1\), which yields \(B = -1\). To find \(C\), we need to choose a value for \(r\) not among the ones we have used. Set \(r = 1\) to get \(C = -1\). So we have: $$ f(r) = \frac{1}{r} - \frac{1}{r + 1} - \frac{1}{r^4 + r^2 + 1}. $$

Evaluate the series

Now, we can rewrite the series as $$ \sum_{r=1}^{\infty} f(r) = \sum_{r=1}^{\infty} \left(\frac{1}{r} - \frac{1}{r + 1} - \frac{1}{r^4 + r^2 + 1}\right), $$ which can be separated into three series as $$ \sum_{r=1}^{\infty} \frac{1}{r} - \sum_{r=1}^{\infty} \frac{1}{r + 1} - \sum_{r=1}^{\infty} \frac{1}{r^4 + r^2 + 1}. $$ Notice that the first two series are telescoping, meaning that most terms will cancel out. Let's see: $$ \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots = 1 - \frac{1}{2}. $$ So the first two series sum up to \(\frac{1}{2}\). For the third series, we can notice that \(r^4 + r^2 + 1 \geq 1\) for all \(r \geq 1\), so the terms in the series are decreasing and positive. Also, \(\frac{1}{r^4 + r^2 + 1} \leq \frac{1}{r^2}\), and the function is integrable on \([1,\infty]\), meaning that the series converges by the integral test. However, it does not contribute to the final answer, as its sum is less than or equal to the integral of \(\frac{1}{x^2}\) from \(1\) to \(\infty\), which is \(1\). Therefore, the value of the given infinite series is equal to $$ \frac{1}{2} + 0\leq \sum_{r=1}^{\infty} f(r) \leq \frac{1}{2} + 1. $$ Among the given answer choices, only (A) \(3 / 2\) is in this range. So, the answer is: (A) \(3 / 2\).