The sum of integer parts of $2k/n$ over $k=1,\dots,n-1$

Question

Let

$$ S_n= \left[\frac{2}{n}\right]+\left[\frac{4}{n}\right]+\cdots+\left[\frac{2(n-1)}{n}\right] $$

where $n$ is an odd integer such that $n\geqslant 3$, and $[x]$ is the integer part of $x$. Show that $$S_n=\frac{n-1}{2}.$$

I thinked about using induction ! But no results.

Answer 1

Notice that $$[a] = a - \{a\}$$ where $\{ a\}$ denotes the fractional part of $a$ Now we may rewrite the sum as $$S_n=\frac{(n+1)*(n+2) }{n}+ \{\frac{2}{n}\}+\{\frac{4}{n}\}+........+\{\frac{2(n+1)}{n}\}$$ Notice that if $a$ is congruent to $c$ modulo $b$ and $c<b$ then $\{\frac a b\} = \frac c b$ If $n$ is odd then $2*i$ is a permutation of all residues $i$ of $n$, thus we may evaluate the above sum as $$S_n=\frac{(n+1)*(n+2) }{n}+ (n-1) + 2 + 2/n = \frac {2*n*(n+1)+2}{n} $$

Answer 2

if $n =2k;$ n is even:

Then $[\frac {2}{n}]$ to $[\frac {n -2}{n}]$ will all be 0. $[\frac {n}{n}]$ to $[\frac {2n -2}{n}]$ will be 1 and that be a total of $\frac n 2$. $[\frac {2n}{n}]$ and $[\frac {2(n+1)}{n}]$ both equal 2 so the sum is $[\frac {n}{2}] +4$.

If $n = 2k + 1;$ n is odd.

Then $[\frac {2}{n}]$ to $[\frac {n -1}{n}]$ will all be 0. $[\frac {n + 1}{n}]$ to $[\frac {2n -2}{n}]$ will be 1 and that be a total of $k = [\frac n 2]$ . $[\frac {2n}{n}]$ and $[\frac {2(n+1)}{n}]$ both equal 2 so the sum is $[\frac {n}{2}] +4$.

Answer 3

You have an indexing error and your statement is simply not true ever because you have added too many terms.

$S_n = \sum_{i = 1}^{n + 1}[\frac {2i}{n}] = \sum_{i = 1; i < \frac n 2}[\frac {2i}{2}] + \sum_{i \ge [\frac n 2]}^{2(n - 1)}[\frac {2i}{n}] + \sum_{i = 2n}^{2(n + 1)}[\frac {2i}{n}]= \sum_{i = 1; i < \frac n 2}0 + \sum_{i \ge [\frac n 2]}^{2(n - 1)}1 + \sum_{i = 2n}^{2(n + 1)}2 = 0 + [\frac{n}{2}] + 2 + 2 = [\frac{n}{2}] + 4$

To get the result you want you want:

$S_n = \sum_{i = 1}^{2(n - 2)}[\frac {2i}{n}] = \sum_{i = 1; i < \frac n 2}[\frac {2i}{2}] + \sum_{i \ge [\frac n 2]}^{2(n - 2)}[\frac {2i}{n}]= \sum_{i = 1; i < \frac n 2}0 + \sum_{i \ge [\frac n 2]}^{2(n - 2)}1= 0 + [\frac{n -1}{2}] = [\frac{n-1}{2}] $

The weird thing is you keep changing restrictions to your equation hoping it will be true for some conditions when it has never been true for any conditions because you have added too many terms.