If $f(r)$ be the integer closest to $\sqrt[4]{r}$, then calculate the value of $\sum^{1995}_{r=1} \frac{1}{f(r)}$

Question

If $f(r)$ be the integer closest to $\sqrt[4]{r}$, then what is the value of $\displaystyle\sum^{1995}_{r=1} \frac{1}{f(r)}$ ?

I am thinking like this way , if we have a number $>1.5$, then its closest integer is $2,$ but I did not understand how to solve that question.

Could some help me to solve it, thanks.

Answer 1

This PARI/GP command calculates the value :

? sum(j=1,1995,1/round(j^(1/4)))
%1 = 400
?

Answer 2

A hint: Use your pocket calculator to produce a table of the values $1.5^4$, $2.5^4$, $\ldots$, $7.5^4$.

Answer 3

Hint: $f$ is a piecewise constant function with jumps from $n$ to $n+1$ at $\lceil (n+0.5)^4\rceil$. You just need to find how long the intervals are in $[1,1995]$.

Surprisingly, the sum is an integer. More surprisingly, each combined term is an integer: $$\sum^{1995}_{r=1} \frac{1}{f(r)} = \frac{5}{1}+\frac{34}{2}+\frac{111}{3}+\frac{260}{4}+\frac{505}{5}+\frac{870}{6}+\frac{210}{7} = 400$$