Polynomial function question

Question

If $f(x)$ is equal to $\frac{1}{x^3 + 3x^2 + x}$, find the smallest value of $n$ for which $f(1) + f(2) + ... F(n) = \frac{503}{2014}$. I tried noting that first initial values of f sum to $\frac{1}{5}$, and $503$ divided by $2014$ is approx $\frac{1}{4}$, so we're aiming for values after $f(2)$ to be equal to $\frac{1}{12}$, we can test values and plug in values for $x$ until we find such a number $n$. This seems algorithmic and I'd try searching for a smarter solution. Is there any other solution? Thanks.

Answer 1

Note that

$$0 < f(x) = \frac{1}{x^3 + 3x^2 + x} \leq \frac{1}{x} + \frac{1}{3x} + \frac{1}{x} = \frac{7}{3x},$$

since $x \in \mathbb{N}$.

Now,

$$f(1) = \frac{1}{5}$$ $$f(2) = \frac{1}{22}$$ $$f(3) = \frac{1}{57}$$

Consequently:

$$f(1) + f(2) = \frac{27}{110} = 0.2\overline{45}$$ $$f(1) + f(2) + f(3) = \frac{1649}{6270} > 0.262998405 > \frac{503}{2014} \approx 0.24975$$

Consequently, there is no such $n \in \mathbb{N}$ such that

$$f(1) + f(2) + \ldots + f(n) = \frac{503}{2014}$$

and

$$f(x) = \frac{1}{x^3 + 3x^2 + x}.$$

Perhaps there is a typo in the definition of the function $f$? =)