Difference between revisions of "2006 iTest Problems/Problem 31"

Problem

The value of the infinite series $$\sum_{n=2}^\infty\dfrac{n^4+n^3+n^2-n+1}{n^6-1}$$ can be expressed as $\tfrac pq$ where $p$ and $q$ are relatively prime positive numbers. Compute $p+q$.

Solution

Notice that $n^6 - 1 = (n^2 - 1)(n^4 + n^2 + 1)$, and notice that the numerator contains $n^4 + n^2 + 1$. Also, notice that $n^3 - n = n(n^2 - 1)$. We can rewrite the expression by factoring and splitting the fractions. $$\sum_{n=2}^\infty\dfrac{n^4+n^2 + 1 + n(n^2 - 1)}{(n^2 - 1)(n^4 + n^2 + 1)}$$ $$\sum_{n=2}^\infty\dfrac{1}{n^2 - 1} + \dfrac{n}{n^4 + n^2 + 1}$$ $$\sum_{n=2}^\infty\dfrac{1}{n^2 - 1} + \dfrac{n}{n^4 + 2n^2 + 1 - n^2}$$ $$\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)} + \dfrac{n}{(n^2 + n + 1)(n^2 - n + 1)}$$ In the fraction $\frac{1}{(n+1)(n-1)}$, we can split the fraction as $\frac{1}{2(n-1)} - \frac{1}{2(n+1)}$, making $\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)}$ a telescoping series. Plugging in values results in $\frac12 - \frac16 + \frac14 - \frac18 + \frac16 - \frac1{10} + \cdots = \frac12 + \frac14 = \frac34$.

In the fraction $\frac{n}{(n^2 + n + 1)(n^2 - n + 1)}$, plugging in the first few values results in $\frac{2}{7 \cdot 3} + \frac{3}{13 \cdot 7} + \frac{4}{21 \cdot 13} + \cdots$. The structure suggests that we could try to model the same fraction with a telescoping series.

Substituting $n$ for $a+1$ in $n^2 - n + 1$ results in $a^2 + a + 1$. Also, because $\tfrac12 \cdot (n^2 + n + 1) - \tfrac12 \cdot (n^2 - n + 1) = n$, we find that the expression is equal to $\frac{1}{2(n^2 - n + 1)} - \frac{1}{2(n^2 + n + 1)}$, confirming that $\sum_{n=2}^\infty\dfrac{n}{(n^2 + n + 1)(n^2 - n + 1)}$ is a telescoping series. The expression equals $\frac{1}{6} - \frac{1}{14} + \frac{1}{14} - \frac{1}{26} + \cdots = \frac{1}{6}$.

Therefore, $\sum_{n=2}^\infty\dfrac{n^4+n^3+n^2-n+1}{n^6-1} = \frac34 + \frac16 = \frac{11}{12}$, so $m+n = \boxed{23}$.