$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$ and other sums of quadratic reciprocals

Question

What is the exact value for $$\sum_{n=1}^{\infty} \frac{1}{n^2+n+1}$$ and can other infinite sums of quadratic reciprocals have specific values.

Answer 1

Infinite Sum $$ \begin{align} \sum_{n=0}^\infty\frac1{n^2+n+1} &=\frac1{i\sqrt3}\sum_{n=0}^\infty\left[\frac1{n+\frac12-i\frac{\sqrt3}2}-\frac1{n+\frac12+i\frac{\sqrt3}2}\right]\tag1\\ &=\frac1{i\sqrt3}\sum_{n=0}^\infty\left[\frac1{n+\frac12-i\frac{\sqrt3}2}+\frac1{-n-\frac12-i\frac{\sqrt3}2}\right]\tag2\\ &=\frac1{i\sqrt3}\sum_{n=-\infty}^\infty\frac1{n+\frac12-i\frac{\sqrt3}2}\tag3\\ &=\frac1{i\sqrt3}\pi\cot\left(\frac\pi2-i\frac{\pi\sqrt3}2\right)\tag4\\ &=\frac1{i\sqrt3}\pi\tan\left(i\frac{\pi\sqrt3}2\right)\tag5\\ &=\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)\tag6 \end{align} $$ Explanation:
$(1)$: partial fractions
$(2)$: rewrite second term
$(3)$: write as a principal value sum
$(4)$: $(7)$ from this answer
$(5)$: $\cot\left(\frac\pi2-x\right)=\tan(x)$
$(6)$: $\tan(ix)=i\tanh(x)$

Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\sum_{n=1}^\infty\frac1{n^2+n+1}=\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)-1}\tag7 $$

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Derivation of Claude Leibovici's Asymptotic Expansion

Since we have $$ \begin{align} \frac1{n^2+n+1} &=\frac1{n^2}\frac{1-\frac1n}{1-\frac1{n^3}}\\ &=\frac1{n^2}-\frac1{n^3}+\frac1{n^5}-\frac1{n^6}+\frac1{n^8}-\frac1{n^9}+\frac1{n^{11}}-\frac1{n^{12}}\cdots\tag8 \end{align} $$ Applying the Euler-Maclaurin Sum Formula to $(8)$ yields $$ \begin{align} &\sum_{k=1}^n\frac1{k^2+k+1}\\ &=\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)-1\\ &-\frac1n+\frac1{n^2}-\frac2{3n^3}+\frac{11}{15n^5}-\frac1{n^6}+\frac1{3n^7}+\frac1{n^8}-\frac8{5n^9}+\frac{83}{33n^{11}}-\frac1{n^{12}}+O\!\left(\frac1{n^{13}}\right)\tag9 \end{align} $$

Answer 2

Too long for a comment.

Starting with limit given by robjohn, it seems that we can get quite good approximation of the partial sums $$S_p=\sum_{n=1}^{p} \frac{1}{n^2+n+1}=\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)-1-\frac{1}{p}+\frac{1}{p^2}-\frac{2}{3 p^3}+\frac{11}{15 p^5}-\frac{1}{p^6}+O\left(\frac{1}{p^7}\right)$$

For $n=10$, the exact result is $\frac{13212635038}{18675445971}\approx 0.7074869890$ while the above expansion gives $\frac\pi{\sqrt3}\tanh\left(\frac{\pi\sqrt3}2\right)-\frac{3271981}{3000000}\approx 0.7074869472$.

Answer 3

According to Maple,

$$ \frac{\pi}{\sqrt{3}} \tanh \left(\frac{\pi \sqrt{3}}{2}\right) - 1 $$

It looks like all the sums $\sum_{n=1}^\infty \frac{1}{n^2+tn+1}$ for integers $t\ne -2 $ have closed forms.

EDIT: Here are a few examples:

$$ \eqalign{\sum_{n=1}^{\infty } \frac{1}{ {n}^{2}+n+1} & =\frac{\pi \tanh(\pi \sqrt{3}/2)}{\sqrt{3}}-1\cr \sum_{n=1}^\infty \frac{1}{n^2+n+1} &= \frac{\pi^2}{6}-1\cr \sum_{n=1}^\infty \frac{1}{n^2+3n+1} &= \frac{\pi \tan(\pi \sqrt{5}/2)}{\sqrt{5}}\cr \sum_{n=1}^\infty \frac{1}{n^2+4n+1} &= -\frac{\pi \cot(\pi \sqrt{3})}{2\sqrt{3}}-\frac{1}{3}\cr \sum_{n=1}^\infty \frac{1}{n^2+5n+1} &= \frac{\pi \tan(\pi \sqrt{21}/2)}{\sqrt{21}}-\frac{7}{15}\cr \sum_{n=1}^\infty \frac{1}{n^2+6n+1} &= -\frac{\pi\cot(2 \sqrt{2}\pi)}{4\sqrt{2}}-\frac{61}{112}}$$