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

[anemone]

Here is this week's POTW:

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

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[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 $$\left\lfloor{\left(n+\dfrac{1}{2}\right)^4}\right\rfloor$$ since

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

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

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

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

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

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