Let $a_n$ and $b_n$ be two positive sequences such that
$$
b_n =\frac{a_1 + a_2 + ... + a_n} {n}
$$
Prove that $\lim_{n\to \infty} a_n = 0$ implies $\lim_{n\to \infty} b_n = 0$. Is the converse true?
Asked
Active
Viewed 97 times
0
Martin Sleziak
- 53,687
1 Answers
1
As Solid Snake pointed it, this is a specific case of
Prove convergence of the sequence $(z1+z2+⋯+zn)/n$ of Cesaro means
I will provide you with an alternate proof for this case assuming we're dealing with real numbers.
Fix $\epsilon>0$. Since $\lim_{n\to\infty}a_n=0$, we can find an $N$ such that for $n>N$, $a_n<\frac{\epsilon}{2}$. Then,
\begin{align} b_n&=\frac{a_1+a_2+...+a_n}{n}\\ &=\frac{a_1+a_2+...+a_N}{n}+\frac{a_{N+1}+...+a_n}{n}\\ &\leq\frac{a_1+a_2+...+a_N}{n}+\frac{(n-N)\frac{\epsilon}{2}}{n}\ \ \ \ (\text{each $a_n<\epsilon$ for $n>N$})\\ &\leq\frac{a_1+a_2+...+a_N}{n}+\frac{\epsilon}{2} \end{align}
Since $N$ is a fixed number, we can find some $K$ such that $a_1+a_2+...+a_N<K$ (that is, a finite sum of of a convergent real sequence can't be infinite). Without loss of generality, we can choose $n>\frac{2K}{\epsilon}$ such that $n>N$ is still true. Continuing from above, we then have
\begin{align} b_n&\leq\frac{a_1+a_2+...+a_N}{n}+\frac{\epsilon}{2}\\ &\leq\frac{K}{\frac{2K}{\epsilon}}+\frac{\epsilon}{2}\\ &\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\\ \end{align}
Thus, for $n$ sufficiently large, $b_n<\epsilon$, showing $\lim_{n\to\infty}b_n=0$. This ends the proof.
The converse is not generally true. Let $a_n=\sin(\frac{\pi}{2}n)$. It is clear that $-1\leq a_1+a_2+...+a_n\leq1$ for any $n$ (i.e. the partial sums are bounded), and thus
$$\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{a_1+a_2+...+a_n}{n}=0.$$
However, $\lim_{n\to\infty}a_n\neq0$.
A few key points to take away:
- This only works for real, positive sequences. If the sequence wasn't positive, we could do some fancy footwork with absolute values to make it work out.
- $\lim_{n\to\infty}b_n=0$ also shows this sequence is Cauchy, since we are dealing with real numbers.
- This is a very specific case of a well-known theorem called Cauchy's First Theorem on Limits.
