Exercise 4.1.20* (Cesàro's mean)

Question

Let $	heta$ be real, and $0 < 	heta < 1$. Define
$$g_n = 
\begin{cases}
  0 & 	ext{if } \lfloor n 	heta floor = \lfloor (n-1) 	hetafloor ,\
  1 & 	ext{otherwise}.
\end{cases} 
$$
Prove that
$$
\lim_{n 	o \infty} \frac{g_1+ g_2 + \cdots + g_n}{n} = 	heta.
$$

Richard Ganaye · Accepted Answer

\begin{proof} 
By definition of the greatest integer function, for any positive integer $n$,
$$\lfloor (n-1) 	heta floor \leq (n-1) 	heta < \lfloor (n-1)	heta floor + 1.$$
Adding $	heta$, where $0 < 	heta < 1$, we obtain
$$ \lfloor (n-1)  	heta floor<  n 	heta < \lfloor (n-1)	heta floor + 2,$$
therefore $\lfloor n 	heta floor = \lfloor (n-1)	heta floor$ or $\lfloor n 	heta floor = \lfloor (n-1) 	heta floor+ 1$.

If $g_n = 1$, then $\lfloor n 	heta floor 
e  \lfloor (n-1)	heta floor$, thus $\lfloor n 	heta floor = \lfloor (n-1) 	heta floor+ 1$. So
$$
\lfloor n 	heta floor -  \lfloor (n-1)	heta floor =
\begin{cases}
0 & 	ext{if } g_n = 0,\
1 & 	ext{if } g_n = 1.
\end{cases}
$$
More concisely, 
$$g_n = \lfloor n 	heta floor -  \lfloor (n-1)	heta floor.$$

Consequently,
\begin{align*}
\sum_{k = 1}^n g_k &= \sum_{k=1}^n ( \lfloor k 	heta floor -  \lfloor (k-1)	heta floor)\
&=  \sum_{k=1}^n  \lfloor k 	heta floor -  \sum_{k=1}^n   \lfloor (k-1)	heta floor\
&= \sum_{j=1}^n  \lfloor j 	heta floor -  \sum_{j=0}^{n-1}   \lfloor j	heta floor & (j = k-1 	ext{ in the second sum}) \
&=  \lfloor n 	heta floor +  \sum_{j=1}^{n-1}   \lfloor j 	heta floor - \sum_{j=1}^{n-1}   \lfloor j 	heta floor  \
&=  \lfloor n 	heta floor.
\end{align*}

Moreover, since $n	heta - 1 < \lfloor n 	heta floor \leq n 	heta$,
$$	heta - \frac{1}{n} < \frac{ \lfloor n 	heta floor}{n} \leq 	heta,$$
where $\lim_{n 	o \infty} \left(	heta - \frac{1}{n} ight)= 	heta$. Therefore
$$\lim_{n 	o \infty} \frac{ \lfloor n 	heta floor}{n} = 	heta.$$

In conclusion, 
$$
\lim_{n 	o \infty} \frac{g_1+ g_2 + \cdots + g_n}{n} = 	heta.
$$
\end{proof}

Exercise 4.1.20* (Cesàro's mean)

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Exercise 4.1.20* (Cesàro's mean)

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