Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$

Question

Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$

$$\begin{align}\frac{1}{n^4+n^2+1}& =\frac{1}{n^4+2n^2+1-n^2}\\ &=\frac{1}{(n^2+1)^2-n^2}\\ &=\frac{1}{(n^2+n+1)(n^2-n+1)}\\ &=\frac{1-n}{2(n^2-n+1)}+\frac{1+n}{2(n^2+n+1)}\\ \end{align}$$

For $n={1,2,3}$ it is not giving telescooping series.

$=\frac{1}{3}+\frac{3}{14}+\frac{2}{13}+0-\frac{1}{6}-\frac{1}{7}$

Answer 1

Partial answer.

Rewrite the terms as:

$$\frac{1}{2}\left(\frac{n+1}{n^2+n+1} - \frac{n-1}{1+(n-1)+(n-1)^2}\right) = \\\frac{1}{2}\left(\frac{1}{n^2+n+1}+f(n)- f(n-1)\right)$$ where $f(n)=\frac{n}{1+n+n^2}$. In particular $\sum_{n=1}^\infty \left(f(n)-f(n-1)\right)=-f(0)=0$.

So we've reduced it to:

$$\frac12\sum \frac{1}{n^2+n+1}$$

Not sure how to compute that sum.

Answer 2

An observation:

We can reduce this series to another one. Note that $$\frac{1}{n(n\pm 1)+1}=\frac{|n\pm 1 -n|}{n(n\pm 1)+1}=|\tan(\arctan (n\pm 1)-\arctan n)|$$ Thus given summation is $$S=\sum_{n=1}^{50}\frac{1}{2n}\left(\tan(\arctan(n)-\arctan(n-1))-\tan(\arctan(n+1)-\arctan n)\right)\\=\sum_{n=1}^{50}\frac{1}{2n}\left(\tan(\beta_n-\beta_{n-1})-\tan(\beta_{n+1}-\beta_n)\right)$$ where $\beta_n:=\arctan n,\ n\ge 1$. I am trying to think if there is a way to find this summation.

Answer 3

Just completing Thomas Andrew's answer, $$\sum_{n=1}^{N}\frac{1}{n^2+n+1}=4\sum_{n=1}^{N}\frac{1}{(2n+1)^2+3}=4\sum_{k=1}^{+\infty}\sum_{k=1}^{N}\frac{3^{k-1}}{(2n+1)^{2k}}\tag{1}$$ can be put in integral form by exploiting: $$ \int_{0}^{+\infty}\frac{\sin(m x)}{m}\,e^{-\sqrt{3}\,x}\,dx = \frac{1}{3+m^2}.\tag{2}$$ That gives, for instance: $$\sum_{n=1}^{+\infty}\frac{1}{n^2+n+1}=4\int_{0}^{+\infty}\left(-\sin x+\varphi(x)\right) e^{-\sqrt{3}\,x}\,dx\\=-1+4\int_{0}^{+\infty}\varphi(x)\,e^{-\sqrt{3}\,x}\,dx \tag{3}$$ where $\varphi(x) = \frac{\pi}{4}\cdot(-1)^{\left\lfloor\frac{x}{\pi}\right\rfloor}$, so that (have also a look at this other question explaining the method): $$\sum_{n=1}^{+\infty}\frac{1}{n^2+n+1}=\color{red}{-1+\frac{\pi}{\sqrt{3}}\tanh\left(\frac{\pi}{2}\sqrt{3}\right)}.\tag{4}$$ By truncating the previous sum at $N=50$ we get something smaller than the RHS of $(4)$, but not much smaller, since the series on the LHS converges quite fast.