Evaluating a Sum Problem: Find Value

[Saitama]

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!

 

[kaliprasad]

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

 

[I like Serena]

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)

 

[Saitama]

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)

 

[CellCoree]

I would approach this problem by first looking for patterns or relationships that can help simplify the calculation. One possible approach is to rewrite the summation as follows:

$$\sum_{m=1}^{\infty} \frac{1}{[\sqrt{m}]^3} = \sum_{m=1}^{\infty} \frac{1}{\lfloor\sqrt{m}\rfloor^3}$$

where $\lfloor\sqrt{m}\rfloor$ represents the nearest integer to $\sqrt{m}$. This allows us to focus on the values of $\sqrt{m}$ rather than the values of $m$ itself.

Next, we can notice that the terms in the summation can be grouped together based on their nearest integer value. For example, all the terms where $\lfloor\sqrt{m}\rfloor = 1$ will be grouped together, followed by all the terms where $\lfloor\sqrt{m}\rfloor = 2$, and so on. This grouping can be represented as follows:

$$\sum_{m=1}^{\infty} \frac{1}{\lfloor\sqrt{m}\rfloor^3} = \left(\frac{1}{1^3} + \frac{1}{1^3} + \frac{1}{1^3} + \dots\right) + \left(\frac{1}{2^3} + \frac{1}{2^3} + \frac{1}{2^3} + \dots\right) + \left(\frac{1}{3^3} + \frac{1}{3^3} + \frac{1}{3^3} + \dots\right) + \dots$$

Now, we can use the geometric series formula to simplify each group:

$$\sum_{m=1}^{\infty} \frac{1}{\lfloor\sqrt{m}\rfloor^3} = \left(\frac{3}{2}\right)^3 + \left(\frac{3}{4}\right)^3 + \left(\frac{3}{6}\right)^3 + \dots$$

Finally, we can use the relationship between $k$ and $2k$ that you noticed to simplify the calculation even further:

$$\sum_{m=1}^{\infty