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How to evaluate following expression? $$ \sum_{n=1}^{\infty}\frac{1}{1+n^2+n^4}$$ I doubt it is a telescopic Sum.

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    $\begingroup$ What do you mean evaluate? This sum has no easily attainable closed form, but it evidently converges by comparing to $a_{n}=n^{-2}.$ To get a sense of the challenges you face in evaluating this sum analytically, see math.stackexchange.com/questions/8337/… for the simpler case $\sum n^{-2}.$ $\endgroup$ – Sargera Oct 10 '14 at 17:22
  • $\begingroup$ Have you attempted partial fraction decomposition? $\endgroup$ – David H Oct 10 '14 at 17:28
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    $\begingroup$ Maple give the answer $$ -1/2+1/6\,\sqrt {3}\pi \,\tanh \left( 1/2\,\pi \,\sqrt {3} \right) $$ $\endgroup$ – Leox Oct 10 '14 at 17:36
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Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then: $$ n^4+n^2+1 = (n^2-\omega)(n^2-\omega^2), $$ so: $$\frac{1}{1+n^2+n^4}=\frac{1}{i\sqrt{3}}\left(\frac{1}{n^2-\omega}-\frac{1}{n^2-\omega^2}\right)$$ and: $$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{\sqrt{3}}\Im\sum_{n=1}^{+\infty}\frac{1}{n^2-\omega}$$ can be computed through the identity: $$\sum_{n=1}^{+\infty}\frac{1}{n^2+a}=\frac{-1+\pi\sqrt{a}\coth(\pi\sqrt{a})}{2a}$$ that follows from considering the logarithmic derivative of the Weierstrass product for the $\sinh$ function. By putting all together, we have:

$$\sum_{n=1}^{+\infty}\frac{1}{1+n^2+n^4}=\frac{1}{6}\left(-3+\pi\sqrt{3}\tanh\frac{\pi\sqrt{3}}{2}\right)$$

as stated by WA.

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First, we have $$ \begin{align} \frac1{z^4+z^2+1} &=\frac1{12}\left( \frac{-3-i\sqrt3}{z-e^{\pi i/3}} +\frac{3+i\sqrt3}{z-e^{4\pi i/3}} +\frac{3-i\sqrt3}{z-e^{2\pi i/3}} +\frac{-3+i\sqrt3}{z-e^{5\pi i/3}} \right)\tag{1} \end{align} $$ Let $\gamma$ be the rectangle $$ [-1-i,1-i]\cup[1-i,1+i]\cup[1+i,-1+i]\cup[-1+i,-1-i]\tag{2} $$ then the integral $$ \frac1{2\pi i}\int_{(n+\frac12)\gamma}\frac{\pi\cot(\pi z)}{z^4+z^2+1}\,\mathrm{d}z\tag{3} $$ tends to $0$ since along the horizontal paths, $|\pi\cot(\pi z)|\to\pi$ and along the vertical paths, $|\pi\cot(\pi z)|\lt\pi$.

Since $\pi\cot(\pi z)$ has residue $1$ at each integer, we get that $(3)$ is the sum of the residues of $(1)$ times $\pi\cot(\pi z)$ at the singularities of $(1)$ plus $$ 1+2\sum_{n=1}^\infty\frac1{n^4+n^2+1}\tag{4} $$ The sum of the residues of $(1)$ times $\pi\cot(\pi z)$ at the singularities of $(1)$ is $$ -4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\cot\left(\pi\tfrac{1+i\sqrt3}2\right)\right)\tag{5} $$ Since the sum of $(4)$ and $(5)$ is $0$, we have $$ \begin{align} 1+2\sum_{n=1}^\infty\frac1{n^4+n^2+1} &=4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\cot\left(\pi\tfrac{1+i\sqrt3}2\right)\right)\\ &=-4\,\mathrm{Re}\left(\tfrac{3+i\sqrt3}{12}\pi\tan\left(\pi\tfrac{i\sqrt3}2\right)\right)\\ &=\tfrac{\pi\sqrt3}3\tanh\left(\pi\tfrac{\sqrt3}2\right)\tag{6} \end{align} $$ Therefore, $$ \sum_{n=1}^\infty\frac1{n^4+n^2+1}=\frac{\pi\sqrt3}6\tanh\left(\pi\frac{\sqrt3}2\right)-\frac12\tag{7} $$

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Let us use the Abel-Plana formula $$ \sum_{n=0}^\infty f(n)= \int_0^\infty f(x) \, dx+ \frac 1 2 f(0)+i \int_0^\infty \frac{f(i t)-f(-i t)}{e^{2\pi t}-1} \, dt $$ for $f(n)=\dfrac{1}{1+n^2+n^4}$.

Since $ f(it)=f(-it) $ we obtain $$ \sum_{n=0}^\infty f(n)= \int_0^\infty f(x) \, dx+ \frac 1 2 f(0). $$ By standard way we get $$ \int_0^\infty\frac{1}{1+x^2+x^4}=\frac{\pi \sqrt{3}}{6}. $$ Now $$ \sum_{i=1}^\infty f(n)=\sum_{i=0}^\infty f(n)-f(0)=\frac{\pi \sqrt{3}}{6}-\frac{1}{2}. $$

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  • $\begingroup$ Have you checked? Jack's answer is approximately $0.39907364028134509045$, and yours is $0.40689968211710892530$. $\endgroup$ – robjohn Oct 11 '14 at 14:09
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    $\begingroup$ You cannot apply the Abel-Plana formula because $f(z) = \frac{1}{1+z^2+z^4}$ is not holomorphic over the left half-plane $\Re z \ge 0$. $f(z)$ has two poles at $\frac12 \pm \frac{\sqrt{3}}{2}i$ there. You can apply AP formula to the sum start at $n = 1$ but then $f(1+it) \ne $f(1-it)$ and the last term in the AP formula won't cancel.... $\endgroup$ – achille hui Oct 11 '14 at 15:04
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    $\begingroup$ Is there any way to fix this problem? $\endgroup$ – Leox Oct 11 '14 at 16:08
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    $\begingroup$ @Leox, the real beauty of your method is $f(it) = f(-it)$ so you don't need to evaluate the last integral which look nasty. In principle, if you can find another sequence which is 1) easy to evaluate, 2) a function in $n^2$, 3) same singular behavior near the two poles, then you can evaluate the difference of this sum and the easier one using this method. However, I have no idea how to get hold of such an easier sequence. $\endgroup$ – achille hui Oct 12 '14 at 0:13
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Hint: Apply residue calculus on $$f(z)= \frac{\pi}{(1+z^{2}+z^{4})\tan(\pi z)}$$ Where you want to integrate over the square $S_{R}$ with vertices $(R+\frac{1}{2})(\pm1 \pm i)$

The reason is that $$Res(f(z), n\pi) = \frac{1}{1+n^{2}+n^{4}}$$ I'll leave the details to you!

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Lots of nice analytical ideas! I will be trying to use them from now one. Numerically, it is not too difficult to get a nice answer. We use the first term of the Euler Maclaurin Summation formula which says:

$$\sum _{n=a}^b f(n)\sim \int_a^b f(x) \, dx+\frac{1}{2} (f(a)+f(b))$$

But it is alot more useful in getting the tail so we first compute

$$ \sum _{n=1}^{99} \frac{1}{n^4+n^2+1}=0.39907330193518 $$

for the integral on the RHS we use the trick that as $x$ gets large the $x^4$ term will completely drown out the $x^2 + 1$ terms.

$$\int_{100}^{\infty } \frac{1}{x^4+x^2+1} \, dx\approx \ \int_{100}^{\infty } \frac{1}{x^4} \, dx=\frac{1}{3000000}$$

Now it is just arithmetic,

$$0.39907330193518\, +\frac{1}{3000000}+\frac{1}{2} \left(\frac{1}{100^4+100^2+1}+0\right)$$

which equals $ 0.39907364026801 $ which is quite close, differing from the exact answer in the 11th place.

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