Find $\left\lfloor\sum{n\over n^4+n^2+1}\right\rfloor-\left\lfloor\sum{n^2\over n^8+n^4+1}\right\rfloor$

Question

I have tried it which goes as follows:

$${n\over n^4+n^2+1}={n\over(n^2+1)^2-n^2}={(n^2+1+n)-(n^2+1-n)\over2(n^2+1-n)(n^2+1+n)}=a_{n-1}-a_n,$$

where $a_n=\dfrac12\dfrac1{n(n+1)+1}$. Thus it becomes a telescopic series:

$$\sum_{n=1}^N{n\over n^4+n^2+1}=a_0-a_N\implies\sum_{n=1}^\infty{n\over n^4+n^2+1}=\lim_{N\to \infty}(a_0-a_N)=a_0-0={1\over2}$$

So

$$\left\lfloor\sum{n\over n^4+n^2+1}\right\rfloor=\left\lfloor{1\over2}\right\rfloor=0$$

But I cannot evaluate the second sum i.e.

$$\left\lfloor\sum{n^2\over n^8+n^4+1}\right\rfloor$$

Can anybody solve the 2nd part? Thanks for assistance in advance.

Answer 1

We don't have to evaluate the second sum. We just need to prove that it's lower than $1$, so that its integer part is $0$. On that note, notice that:

$$\frac{n^2}{n^8+n^4+1} \leq \frac{n^2}{3n^4} = \frac{1}{3}\cdot \frac{1}{n^2} \leq \frac{1}{3} \cdot \frac{1}{n(n-1)} = \frac{1}{3}\left( \frac{1}{n-1}-\frac{1}{n}\right)$$

for any $n \geq 2$. So:

$$\sum_{n=1}^{\infty} \frac{n^2}{n^8+n^4+1} = \frac{1}{3} +\sum_{n=2}^{\infty} \frac{n^2}{n^8+n^4+1}\leq \frac{1}{3}+\frac{1}{3}\sum_{n=2}^{\infty} \left( \frac{1}{n-1}-\frac{1}{n}\right) = \frac{1}{3}+\frac{1}{3}=\frac{2}{3}$$

In conclusion

$$\left\lfloor\sum_{n=1}^{\infty} \frac{n^2}{n^8+n^4+1}\right\rfloor = 0$$

and your final answer is $0$ as well.