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

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    2017
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The discussion centers on evaluating the sum of reciprocals, specifically \sum_{i=1}^{1995}\dfrac{1}{f(i)}, where f(k) represents the integer closest to the fourth root of k. Participants are encouraged to engage with the Problem of the Week (POTW) and follow the provided guidelines for submissions. There is a note expressing disappointment that last week's problem went unanswered. A suggested solution is available for those interested in the methodology. Engaging with these mathematical challenges can enhance problem-solving skills.
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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!
 
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No one answered last week's problem.(Sadface)

You can find the suggested solution below:

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

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