One of friend gave me a question today to solve which is as follows

$$\sum_{r=1}^{\infty} \sqrt {\frac {r}{r^4+r^2+1}}$$

In spite of much efforts I couldn't solve it and so I asked him to check whether the question was correct or was it this one

$$\sum_{r=1}^{\infty} \frac {r}{r^4+r^2+1}$$

I thought the question would be this one because the terms inside the root can be telescoped in absence of root. And indeed I was right. The question was as I expected the latter one.

But even after that I thought about whether the first question containing the square root could also be solved or not. For just checking out the convergence of the sequence I tried the ratio test but it wasn't quite helpful. Then I tried using the integral test and indeed $$\lim_{t\to \infty} \int_{1}^{t} \sqrt{\frac {x}{x^4+x^2+1}} dx$$

this integral converges to $1.80984$ according to Wolfy. Upon lot of efforts too I am not an able to solve the first summation (with the square root terms) . Can someone please lend me some help over this problem.

  • $\begingroup$ You can try to rationalize the denominator to get rid of the square root. Then use partial fractions... $\endgroup$
    – Frank W
    Jun 5, 2018 at 15:42
  • $\begingroup$ @Frank W. What should I take the rationalizing factor as? Cause I don't see anything popping up $\endgroup$ Jun 5, 2018 at 15:44
  • $\begingroup$ Yeah, I don't think rationalizing the denominator is gonna work here. @FrankW. You probably aren't going to be able to get a telescoping series because partial sums are a bit mad, and telescoping sequences have nice partial sums. $\endgroup$ Jun 5, 2018 at 15:55
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    $\begingroup$ @ThomasAndrews Yes, I tried that route. Just went uglier and uglier. If you enter it into Wolfram Alpha, you get a decimal approximation. So most likely, there isn't quite a closed form for this sum. $\endgroup$
    – Frank W
    Jun 5, 2018 at 16:13
  • $\begingroup$ @Manthanein I was thinking that you could multiply the denominator and numerator by $(r^4+r^2+1)^{1/2}$ so the denominator is free of any square roots. Didn't work though, the result was still rather ugly. $\endgroup$
    – Frank W
    Jun 5, 2018 at 16:13

1 Answer 1


We have $$\sqrt{\frac{r}{r^4+r^2+1}}\approx \frac{2}{\sqrt{r-1/2}}-\frac{2}{\sqrt{r+1/2}},$$ so by creative telescoping the value of the series is not too far from $\frac{1+2\sqrt{2}}{\sqrt{3}}$, and this approach can be improved in order to obtain more accurate approximations. However I won't bet on a simple closed form for the given series. Numerical methods produce $\approx 2.12534074896$, which not by chance is pretty close to $$ \int_{1/2}^{+\infty}\sqrt{\frac{x}{x^4+x^2+1}}\,dx\approx 2.12, $$ an incomplete elliptic integral.


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