Can You Solve the Sum of Reciprocals for the Fourth Root Function?

[anemone]

Here is this week's POTW:

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Evaluate \(\displaystyle \sum_{i=1}^{1995}\dfrac{1}{f(i)}\), given that \(\displaystyle f(k)\) be the integer closest to \(\displaystyle \sqrt[4]{k}\).

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[anemone]

No one answered last week's problem.(Sadface)

You can find the suggested solution below:

Spoiler

The largest integer with fourth root closest to $n$ is \(\displaystyle \left\lfloor{\left(n+\dfrac{1}{2}\right)^4}\right\rfloor\) since

\(\displaystyle \begin{align*}\left\lfloor{\left(n+\dfrac{1}{2}\right)^4}\right\rfloor&=\left\lfloor{n^4+2n^3+\dfrac{3}{2}n^2+\dfrac{1}{2}n+\dfrac{1}{16}}\right\rfloor\\&=\left\lfloor{n^4+2n^3+\dfrac{1}{2}\left(3n^2+k\right)+\dfrac{1}{16}}\right\rfloor\\&=n^4+2n^3+\dfrac{1}{2}\left(3n^2+n\right)\text{since } 3n^2+n\text{ is even}\end{align*}\)

$\therefore$ the number of integers with fourth root closest to $n$ is:

\(\displaystyle \begin{align*}\left\lfloor{\left(n+\dfrac{1}{2}\right)^4}\right\rfloor-\left\lfloor{\left((n-1)+\dfrac{1}{2}\right)^4}\right\rfloor&=\left\lfloor{\left(n+\dfrac{1}{2}\right)^4}\right\rfloor-\left\lfloor{\left(n-\dfrac{1}{2}\right)^4}\right\rfloor\\&=\left(n^4+2n^3+\dfrac{1}{2}\left(3n^2+n\right)\right)-\left(n^4-2n^3+\dfrac{1}{2}\left(3n^2-n\right)\right)\\&=4n^3+n\end{align*}\)

That is, $f(k)=n$ for $4n^3+n$ (consecutive) values of $k$.

Since \(\displaystyle f(1995)=7,\,\sum_{n=1}^{6}(4n^3+n)=1785\) and $f(1786)=f(1787)=\cdots=f(1995)=7$, it follows that

\(\displaystyle \begin{align*}\sum_{i=1}^{1995}\dfrac{1}{f(i)}&=\sum_{i=1}^{1785}\dfrac{1}{f(i)}+\dfrac{210}{7}\\&=\sum_{n=1}^{6}\left(\frac{4n^3+n}{n}\right)+30\\&=\sum_{n=1}^{6}\left(4n^2+1\right)+30\\&=400\end{align*}\)