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$$ S = \sum_{k=1}^{n} \frac{k}{k^{4} + k^{2} + 1} $$

I started by factorizing the denominator as $k^2+k+1$ and $k^2-k+1$ The numerator leaves a quadratic with $k$ and $k-1$ or a constant with $k+1$ and $k-1.$ I tried writing the individual terms, ofcourse, it was useless. How do I do this?

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    $\begingroup$ After you factor the denominator and use partial fractions, I believe the sum will telescope. $\endgroup$ – User8128 Nov 30 '17 at 14:41
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    $\begingroup$ As a rule, we don't use $\sum_{n=1}^{n}$. We use two different variables: $\sum_{k=1}^{n}\dots$. Also, you *really don't need any of those parantheses. $\endgroup$ – Thomas Andrews Nov 30 '17 at 14:44
  • $\begingroup$ Related: math.stackexchange.com/questions/2449134/… $\endgroup$ – Jack D'Aurizio Nov 30 '17 at 17:30
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So your term is equal to $$\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{k^2+k+1}\right)$$

Now note $(k+1)^2-(k+1)+1=k^2+k+1$, so your term is: $$\frac{1}{2}\left(\frac{1}{k^2-k+1}-\frac{1}{(k+1)^2-(k+1)+1}\right)$$

and you can apply a telescoping series technique to establish the sum to $n$ is just half of $\frac{1}{1^2 -1 +1}-\frac{1}{(n+1)^2-(n+1)+1}$. And if you are also looking for the limit, half of $\frac{1}{1^2 -1 +1}$.

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  • $\begingroup$ Note that if you didn't recognize that $(k+1)^2-(k+1)+1=k^2+k+1$, you can discover it by noticing that the roots of $x^2-x+1=0$ - namely, $\frac12(1\pm\sqrt{-3})$ - are exactly 1 more than the roots of $x^2+x+1=0$, $\frac12(-1\pm\sqrt{-3})$. $\endgroup$ – Steven Stadnicki Dec 1 '17 at 17:06

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