MHB Evaluating a Sum Problem: Find Value

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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.
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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!
 
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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)
 
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