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.
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Ah, you're right. My bad. – user164403 Nov 06 '14 at 23:50
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1Factor 2014 and use the fact that when adding up fractions, prime numbers in denominators can vanish but they can't appear out of thin air. – Count Iblis Nov 07 '14 at 00:08
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Hey @user164403, please do check out my answer below! =) – Jose Arnaldo Bebita Dris Nov 07 '14 at 00:58
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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 $f$? =)
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Ah, that was a good answer. There was a typo, this is an inequality not an equality haha. Thank you! – user164403 Nov 16 '14 at 01:56
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@user164403, if you think this is a good answer, please consider upvoting and choosing it as the best answer! =) – Jose Arnaldo Bebita Dris Nov 16 '14 at 13:56