MHB Evaluating a Sum Problem: Find Value

  • Thread starter Thread starter Saitama
  • Start date Start date
  • Tags Tags
    Sum
AI Thread Summary
The discussion focuses on evaluating the infinite sum $$\sum_{m=1}^{\infty} \frac{1}{[\sqrt{m}]^3}$$ where $[x]$ denotes the nearest integer to $x$. Participants analyze how often each integer $k$ appears in the sequence generated by $\sqrt{m}$, concluding that each integer $k$ is repeated $2k$ times. This is derived from examining the intervals around $m^2$ that correspond to each integer. The conversation emphasizes the importance of understanding the distribution of integers in the sum to simplify calculations. Overall, the insights provided help clarify the approach to solving the sum problem.
Saitama
Messages
4,244
Reaction score
93
Problem:
Let $[x]$ be the nearest integer to $x$. (For $x=n+\frac{1}{2}, n\in \mathbb{N}$, let $[x]=n$).

Find the value of
$$\sum_{m=1}^{\infty} \frac{1}{[\sqrt{m}]^3}$$

Attempt:

I tried writing down a few terms and saw that $1$ repeats $2$ times, $2$ repeats $4$ times but I didn't check it for three. I think $k$ repeats $2k$ times but is there way to come to this conclusion without checking a few initial numbers and avoid the laborious calculation?

Any help is appreciated. Thanks!
 
Mathematics news on Phys.org
Pranav said:
Problem:
Let $[x]$ be the nearest integer to $x$. (For $x=n+\frac{1}{2}, n\in \mathbb{N}$, let $[x]=n$).

Find the value of
$$\sum_{m=1}^{\infty} \frac{1}{[\sqrt{m}]^3}$$

Attempt:

I tried writing down a few terms and saw that $1$ repeats $2$ times, $2$ repeats $4$ times but I didn't check it for three. I think $k$ repeats $2k$ times but is there way to come to this conclusion without checking a few initial numbers and avoid the laborious calculation?

Any help is appreciated. Thanks!

do you want is nearest to $\sqrt{x}$

then let us look at the numbers nearest to $m^2$ they are from $m^2-(m-1)$ to $m^2+m$ that is m occurs 2m times
 
Hey Pranav! ;)

Let's see how often any specific number $k$ is repeated.If we would have $\sqrt m =k+\frac 1 2$, the number $k$ would just be repeated a last time.
That is, when we would have $m=(k+\frac 1 2)^2 = k^2+k+\frac 1 4$.
Since $m$ is an integer, $k$ is repeated for the last time when $m=k^2+k$.Consequently, $k-1$ was last repeated when $m=(k-1)^2+(k-1)$.
In other words, the number $k$ is repeated $\Big(k^2+k\Big) - \Big((k-1)^2+(k-1)\Big) = 2k$ times.Yes! You were right! (Sun)
 
I like Serena said:
Hey Pranav! ;)

Let's see how often any specific number $k$ is repeated.If we would have $\sqrt m =k+\frac 1 2$, the number $k$ would just be repeated a last time.
That is, when we would have $m=(k+\frac 1 2)^2 = k^2+k+\frac 1 4$.
Since $m$ is an integer, $k$ is repeated for the last time when $m=k^2+k$.Consequently, $k-1$ was last repeated when $m=(k-1)^2+(k-1)$.
In other words, the number $k$ is repeated $\Big(k^2+k\Big) - \Big((k-1)^2+(k-1)\Big) = 2k$ times.Yes! You were right! (Sun)

That was really nicely presented and explained. Thanks a lot ILS! :) (Sun)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top