An Identity From the 2010 China Mathematical Competition (for high schools)

The following problem has been offered at the 2010 China Mathematical Competition for high schools [Lecture Notes, p 30]:

Let $n$ be any natural number. Prove the identity

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

It is curious why the authors of the problem decided to use the "sum form" $\displaystyle\sum_{k=1}^{n}k$ instead of the well known total $\displaystyle\frac{n(n+1)}{2}$.

Solution

References

1. Xu Jiagu, Lecture Notes on Mathematical Olympiad Courses, v 8, (For senior section, v 2), World Scientific, 2012

Let $n$ be any natural number. Prove the identity

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

Solution 1

Let's replace the sum on the right with the closed form: $\displaystyle\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$ and see whether we may apply mathematical induction to prove the resulting identity:

$\displaystyle \sum_{k=1}^{n}\frac{k}{k^{4}+k^{2}+1}=\frac{n(n+1)}{2(n^{2}+n+1)}.$

For $n=1$ both sides give $\displaystyle\frac{1}{3}$. So, let's assume the identity holds for $n=k$, and prove it for $n=k+1$. Denoting the right-had side $\displaystyle f(n)=\frac{n(n+1)}{2(n^{2}+n+1)}$, we have for the difference

\displaystyle \begin{align} f(k+1)-f(k) &= \frac{(k+1)(k+2}{2((k+1)^{2}+(k+1)+1)} - \frac{k(k+1)}{2(k^{2}+k+1)} \\ &= \frac{k+1}{2}\bigg[\frac{(k+1)+1}{(k+1)^{2}+(k+1)+1} - \frac{(k+1)-1}{(k+1)^{2}-(k+1)+1}\bigg] \\ &= \frac{k+1}{2}\frac{[(k+1)^{3}+1] - [(k+1)^{3}-1]}{((k+1)^{2}+(k+1)+1)((k+1)^{2}-(k+1)+1)} \\ &= \frac{k+1}{[(k+1)^{2}+1]^{2}-(k+1)^{2}} \\ &= \frac{k+1}{(k+1)^{4}+(k+1)^{2}+1}, \\ \end{align}

as expected.

Solution 2

Here we reduce the sum on the left to a somewhat unusual telescoping series:

\displaystyle \begin{align} \sum_{k=1}^{n}\frac{k}{k^{4}+k^{2}+1} &= \sum_{k=1}^{n}\frac{k}{(k^{2}+1)^{2}-k^{2}} \\ &= \sum_{k=1}^{n}\frac{k}{(k^{2}+1+k)(k^{2}+1-k)} \\ &= \frac{1}{2}\sum_{k=1}^{n}\bigg[\frac{1}{k^{2}+1-k}-\frac{1}{k^{2}+1+k}\bigg] \\ &= \frac{1}{2}\bigg[1-\frac{1}{n^{2}+1+n}\bigg] \\ &= \frac{1}{2}\frac{n^{2}+n}{n^{2}+1+n} \\ &= \frac{1}{n^{2}+n+1}\frac{n^{2}+n}{2} \\ &= \frac{1}{n^{2}+n+1}\sum_{k=1}^{n}k. \end{align}