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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.

DXT
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    there is no equation... actually there is no question. – Surb Feb 28 '18 at 13:01
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    I think he means find the sum.. – King Tut Feb 28 '18 at 13:01
  • What is the source of the problem, and why is it of mathematical interest ? The unmotivated number '1995' suggests this might just be some kind of contest problem. Why was that number chosen rather than any other? – Carl Mummert Mar 01 '18 at 13:21

3 Answers3

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This PARI/GP command calculates the value :

? sum(j=1,1995,1/round(j^(1/4)))
%1 = 400
?
Peter
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  • You lazy boy. One can do it by hand ; to see why, try vector(1995,j,1/round(j^(1/4))) in GP and you'll see – Ewan Delanoy Feb 28 '18 at 13:20
  • True, on the other hand, the author did not react to the question whether he/she wants a solution by hand ... – Peter Feb 28 '18 at 13:21
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    Indeed. I hate those posts where the author is unresponsive ... Hopefully he'll react soon – Ewan Delanoy Feb 28 '18 at 13:23
  • The main effort is to find the exact bounds for the numbers falling into the different classes given by $f(n)=c$ – Peter Feb 28 '18 at 13:23
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    For the relatively small number 1995, this is doable by hand, even if tedious. I agree that one should not make that effort for an asker who isn't even minimally responsive – Ewan Delanoy Feb 28 '18 at 13:26
  • Well, should I add a hint to encourage the author to solve it him/herself ? – Peter Feb 28 '18 at 13:34
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    It's already all there in the comments we just made here, I think ... – Ewan Delanoy Feb 28 '18 at 13:36
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A hint: Use your pocket calculator to produce a table of the values $1.5^4$, $2.5^4$, $\ldots$, $7.5^4$.

Christian Blatter
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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: $$\sum^{1995}_{r=1} \frac{1}{f(r)} = \frac{5}{1}+\frac{34}{2}+\frac{111}{3}+\frac{260}{4}+\frac{505}{5}+\frac{870}{6}+\frac{210}{7} = 400$$

lhf
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