Number Example 28 (2005 National High School Mathematics Competition Question) For each positive integer $n$, define the function
interval
question
$f(n)=\left\{\begin{array}{ll}0, & \text { if } n \text { is a perfect square, } \\ {\left[\frac{1}{\{\sqrt{n}\}}\right] ; \text { if } n \text { is not a perfect square. }}\end{array}\right.$
(Here $[x]$ denotes the greatest integer not exceeding $x$, and $\{x\}=x-[x]$). Try to find: $\sum_{k=1}^{240} f(k)$.

Solution 1 For any $a, k \in \mathbf{N}_{+}$, if $k^{2}<a<(k+1)^{2}$, let $a=k^{2}+m, m=1,2, \cdots, 2 k$,
$\sqrt{a}=k+\theta, \quad 0<\theta<1$,
then
$\left[\frac{1}{\{\sqrt{a}\}}\right]=\left[\frac{1}{\sqrt{a}-k}\right]=\left[\frac{\sqrt{a}+k}{a-k^{2}}\right]=\left[\frac{2 k+\theta}{m}\right]$.
Since
$0<\frac{2 k+\theta}{m}-\frac{2 k}{m}<1$,
if there exists an integer $t$ between $\frac{2 k}{m}$ and $\frac{2 k+\theta}{m}$, then $\frac{2 k}{m}<t<\frac{2 k+\theta}{m}$.
Thus, on one hand, $2 k<m t$, so $2 k+1 \leqslant m t$, on the other hand, $m t<2 k+\theta<2 k+1$, which is a contradiction. Therefore, $\left[\frac{2 k+\theta}{m}\right]=\left[\frac{2 k}{m}\right]$,
so $\sum_{k^{2}<a<(k+1)^{2}}\left[\frac{1}{\{a\}}\right]=\sum_{m=1}^{2 k}\left[\frac{2 k}{m}\right]$,
thus $\sum_{u=1}^{(n+1)^{2}} f(a)=\sum_{k=1}^{n} \sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$.
Next, we calculate $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$: Draw a $2 k \times 2 k$ table, in the $i$-th row, fill in “*” at the multiples of $i$,
then the number of “*” in this row is $\left[\frac{2 k}{i}\right]$, the total number of “*” in the entire table is $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]$; on the other hand, collect the number of “*” by column:
In the $j$-th column, if $j$ has $T(j)$ positive divisors, then this column has $T(j)$ “*”, so the total number of “*” in the entire table
is $\sum_{j=1}^{2 k} T(j)$, therefore $\sum_{i=1}^{2 k}\left[\frac{2 k}{i}\right]=\sum_{j=1}^{2 k} T(j)$.
Example as follows:
Then
$\sum_{a=1}^{(n+1)^{2}} f(a)=\sum_{k=1}^{n} \sum_{j=1}^{2 k} T(j)$
$$
=\sum_{k=1}^{n}[T(1)+T(2)]+(n-1)[T(3)+T(4)]+\cdots+
$$
$[T(2 n-1)+T(2 n)]$.
Thus $\sum_{k=1}^{16^{2}} f(k)=\sum_{k=1}^{15}(16-k)[T(2 k-1)+T(2 k)]$.
(2)
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Let $a_{k}=T(2 k-1)+T(2 k), k=1,2, \cdots, 15$, it is easy to get the values of $a_{k}$ as follows:
Therefore $\sum_{k=1}^{256} f(k)=\sum_{k=1}^{15}(16-k) a_{k}=783$.
By definition $f(256)=f\left(16^{2}\right)=0$, when $k \in\{241,242, \cdots, 255\}$, let $k=15^{2}+r(16 \leqslant r \leqslant 30)$,
$\sqrt{k}-15=\sqrt{15^{2}+r}-15=\frac{r}{\sqrt{15^{2}+r}+15}$,
$\frac{r}{31}<\frac{r}{\sqrt{15^{2}+r}+15}<\frac{r}{30}$,
$1 \leqslant \frac{30}{r}<\frac{1}{\left\{\sqrt{15^{2}+r}\right\}}<\frac{31}{r}<2$,
\begin{tabular}{c}
\hline A \\
Forbidden \\
C \\
Number \\
Insect \\
of \\
Numbers \\
Interval \\
Question
\end{tabular}
then $\left[\frac{1}{\{\sqrt{k}\}}\right]=1, k \in\{241,242, \cdots, 255\}$,
thus $\sum_{k=1}^{240} f(k)=783-\sum_{k=1}^{256} f(k)=783-15=768$.
Solution 2 Since $15^{2}<240<16^{2}$.
Since $f(k)=0$ when $k$ is a perfect square, we have
$\sum_{k=1}^{240} f(k)=\sum_{k=1+1}^{2^{2}-1} f(k)+\sum_{k=2^{2}+1}^{3^{2}-1} f(k)+\cdots+\sum_{k=14^{2}+1}^{15^{2}-1} f(k)+\sum_{k=226}^{240} f(k)$
$$
=\sum_{n=1}^{14} \sum_{k=n^{2}+1}^{(n+1)^{2}-1} f(k)+\sum_{k=226}^{240} f(k) \text {. }
$$
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Dong
Lan
This
1
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Start
When $n^{2}+1 \leqslant k \leqslant(n+1)^{2}-1$, $[\sqrt{k}]=n,\{k\}=\sqrt{k}-[k]=\sqrt{k}-n$.
Hence $\left[\frac{1}{\{\sqrt{k}\}}\right]=\left[\frac{1}{\sqrt{k}-n}\right]=\left[\frac{\sqrt{k}+n}{k-n^{2}}\right]=\left[\frac{[\sqrt{k}+n]}{k-n^{2}}\right]=\left[\frac{2 n}{k-n^{2}}\right]$.
Therefore $\sum_{k=n^{2}+1}^{(n+1)^{2}-1} f(k)=\sum_{k=n^{2}+1}^{(n+1)^{2}-1}\left[\frac{2 n}{k-n^{2}}\right]=\sum_{i=1}^{2 n}\left[\frac{2 n}{i}\right]$,
$$
\begin{array}{l}
\sum_{k=226}^{240} f(k)=\sum_{k=226}^{240}\left[\frac{2 \cdot 15}{k-15^{2}}\right]=\sum_{i=1}^{15}\left[\frac{30}{i}\right] \text {. } \\
\text { Hence } \sum_{k=1}^{240} f(k)=\sum_{n=1}^{14} \sum_{i=1}^{2 n}\left[\frac{2 n}{i}\right]+\sum_{i=1}^{15}\left[\frac{30}{i}\right] \\
=\sum_{n=1}^{14}\left(\sum_{i=1}^{n}\left[\frac{2 n}{i}\right]+\sum_{i=n+1}^{2 n}\left[\frac{2