For any integer $n$ define $k(n)=\frac {n^7} 7+\frac{n^3}{3}+\frac {11n}{21}+1$ and $$f(n)=\begin{cases} 0 & \text{if $k(n)$ is an integer} \\ 1/n^2 & \text{if $k(n)$ is not an an integer}\end{cases}$$

Find, $$\sum_{n=-\infty}^\infty f(n)$$

My attempt

I hope that if we guessed for what values of $n$ $k(n)$ will be integer. If i have that I could express this series interms of $\sum \frac 1 {n^2}$. And so, we can find the sum.

If no $n$ provides interger for $k(n)$ then its easy. the sum $=\dfrac{\pi^2}6$. Help me to do so.


1 Answer 1


$$k(n)=\frac {n^7} 7+\frac{n^3}{3}+\frac {11n}{21}+1 = \frac {n^7-n} 7+\frac{n^3 - n}{3}+\left(\frac {11}{21}+\frac{1}{7}+\frac{1}{3}\right)n+1$$

Form Fermat's little theorem $p|n^p - n$ for any prime $p$, hence $k(n)$ is always an integer.


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