  Q: Reducing $\sum_{n=1}^\infty \frac{1}{n^4+n}$ using contour integration. Using contour integration, the sum $$\sum_{n=1}^\infty \frac{1}{n^4+n}$$ can be calculated to be $-\frac{3 \pi ^3}{4 \times 32}$. I have proven this with the substitution $x= \tan(\theta/2)$. However, I have also been asked to use contour integration, and I have never used it before. What would be the best way to go about calculating this sum using contour integration? A: Let $f(z) = 1/(z+1)$, $U = \mathbb{C} \setminus [1,\infty)$, $V = (-\infty,1)$ and $R$ be any contour around the real line connecting $V$ and $U$. Using the Residue Theorem, we get that $$\sum_{n=1}^{\infty} \frac{1}{n^4+n} = \frac{1}{2\pi i} \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty} \frac{1}{x^3(x+1)} \, dx.$$ Notice that $f(z)$ is analytic and vanishes on the contour $R$. Thus, the only singularity inside the contour is at $x=0$, and since $x+1$ is analytic at $x=0$, we can apply Cauchy’s Integral Theorem, which says that $$\frac{1}{2\pi i} \int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty} \frac{1}{x^3(x+1)} \, dx = \frac{1}{2\pi i} \int_{\gamma} \frac{1}{x^3(x+1)} \, dx.$$ This is an $n$-fold integral of a function $h(x)$ analytic in $U$ and vanishes on

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