Show that the limit is equal to zero

mathmari

Well-known member
MHB Site Helper
It is given that $a_{n}$ is a positive and decreasing sequence.
Show that if $\sum_{n=1}^{\infty}a_{n}$ converges, $\lim_{n \to \infty}na_{n}=0$.
That's what I tried.Could you tell me if it is right??
$\sum_{n=1}^{\infty}a_{n}$ converges,so it satisfies the Cauchy criterion,so :
$\forall m,n$ with $m>n$ it exists a $n_{0}$ such that for $m,n \geq n_{0}$:$|a_{n}+a_{n+1}+....+a_{m-1}+a_{m}|< \epsilon$ .From the Triangle inequality,we have that $|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}|< \epsilon$ and because of the fact that $a_{n}$ is decreasing,$|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}| \leq |a_{n}|+|a_{n}|+....+|a_{n}|+|a_{n}|=n|a_{n}|=|na_{n}| < \epsilon$.
$$|na_{n}| < \epsilon$$ ,so $\lim_{n \to \infty}na_{n}=0$

chisigma

Well-known member
It is given that $a_{n}$ is a positive and decreasing sequence.
Show that if $\sum_{n=1}^{\infty}a_{n}$ converges, $\lim_{n \to \infty}na_{n}=0$.
That's what I tried.Could you tell me if it is right??
$\sum_{n=1}^{\infty}a_{n}$ converges,so it satisfies the Cauchy criterion,so :
$\forall m,n$ with $m>n$ it exists a $n_{0}$ such that for $m,n \geq n_{0}$:$|a_{n}+a_{n+1}+....+a_{m-1}+a_{m}|< \epsilon$ .From the Triangle inequality,we have that $|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}|< \epsilon$ and because of the fact that $a_{n}$ is decreasing,$|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}| \leq |a_{n}|+|a_{n}|+....+|a_{n}|+|a_{n}|=n|a_{n}|=|na_{n}| < \epsilon$.
$$|na_{n}| < \epsilon$$ ,so $\lim_{n \to \infty}na_{n}=0$
If the series $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges, then for any h>0 there is an $n_{0}$ for which for all $n>n_{0}$ is...

$\displaystyle a_{n} < \frac{h}{n} \implies n\ a_{n} < h\ (1)$

... and that means that $\displaystyle \lim_{n \rightarrow \infty} n\ a_{n} = 0$...

Kind regards

$\chi$ $\sigma$

mathmari

Well-known member
MHB Site Helper
If the series $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges, then for any h>0 there is an $n_{0}$ for which for all $n>n_{0}$ is...

$\displaystyle a_{n} < \frac{h}{n} \implies n\ a_{n} < h\ (1)$

... and that means that $\displaystyle \lim_{n \rightarrow \infty} n\ a_{n} = 0$...

Kind regards

$\chi$ $\sigma$
And if I have to use the Cauchy Criterion for series....how can I solve this?

Evgeny.Makarov

Well-known member
MHB Math Scholar
If the series $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges, then for any h>0 there is an $n_{0}$ for which for all $n>n_{0}$ is...

$\displaystyle a_{n} < \frac{h}{n}$
I think the challenge is to prove this.

$\sum_{n=1}^{\infty}a_{n}$ converges,so it satisfies the Cauchy criterion,so :
$\forall m,n$ with $m>n$ it exists a $n_{0}$ such that for $m,n \geq n_{0}$:$|a_{n}+a_{n+1}+....+a_{m-1}+a_{m}|< \epsilon$. .From the Triangle inequality,we have that $|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}|< \epsilon$.
First, in the Cauchy criterion $m$ and $n$ should be quantified only once: For a given $\epsilon>0$ there exists an $n_0$ such that for all $n_0\le n\le m$... Second, the last inequality holds not by the triangle inequality but because
$|a_{n}+a_{n+1}+....+a_{m-1}+a_{m}|=a_{n}+a_{n+1}+....+a_{m-1}+a_{m}$
since all $a_i$ are positive by assumption. If it were not so, we could only conclude that
$|a_{n}+a_{n+1}+....+a_{m-1}+a_{m}|\le|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}|$
and the fact that the left-hand side is $<\epsilon$ does not imply that so is the right-hand side.

and because of the fact that $a_{n}$ is decreasing,$|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}| \leq |a_{n}|+|a_{n}|+....+|a_{n}|+|a_{n}|=n|a_{n}|=|na_{n}| < \epsilon$.
Here, the right-hand side of the last equality should be $(m-n+1)a_n$ and not $na_n$ since
$|a_{n}|+|a_{n+1}|+....+|a_{m-1}|+|a_{m}|$
has $m-n+1$ terms. But we can take $m=2n-1$ so that $m-n+1=n$. More importantly, the two facts
\begin{align}
a_n+\dots+a_m&\le na_n&&(1)\\
a_n+\dots+a_m&<\epsilon&&(2)\\
\end{align}
do not imply that $na_n<\epsilon$. For this, the inequality in (1) should be reversed.

Suppose that for a given $\epsilon$ we found an $n_0$ such that for all $m$ and $n$,
$n_0\le n\le m\implies a_n+\dots+a_m<\epsilon\qquad(3)$
We would like to find something smaller than $a_n+\dots+a_m$ to conclude from (3) that it is $<\epsilon$. Therefore, it makes sense to compare each term in $a_n+\dots+a_m$ not with $a_n$, but with $a_m$:
$(m-n+1)a_m\le a_n+\dots+a_m$
But we would like to have something like $ma_m$ and not $(m-n+1)a_m$. We can't have $ma_m$ because we don't have enough terms: we can start only from $a_n$ and not from $a_1$. Alternatively, we could put $m=2n$; then, roughly speaking, there are $n$ terms of whose the last is $a_{2n}$, so $na_{2n}\le a_n+\dots+a_m<\epsilon$. But to show that $na_n\to0$ we need to show that every $na_n$ (starting from some point) is small; it would not suffice to show this only for a subsequence $a_{2n}$. Note, however, that $na_n\to0$ is equivalent to $(n/2)a_n\to0$. So the idea is to consider sums from $n/2$ to $n$ (roughly speaking because one needs to take care of odd $n$ for which $n/2$ is not defined) and to compare such sums with $n/2$ times its smallest term, i.e., last term $a_n$. Then we would get $(n/2)a_n\le a_{n/2}+\dots+a_n<\epsilon$.

chisigma

Well-known member
I think the challenge is to prove this...
If they exist an $h > 0$ and an $n_{0}$ such that for any $n>n_{0}$ is $a_{n} \ge \frac{h}{n}$, then the series diverges and that is the prove...

Kind regards

$\chi$ $\sigma$

Evgeny.Makarov

Well-known member
MHB Math Scholar
If the series $\displaystyle \sum_{n=1}^{\infty} a_{n}$ converges, then for any h>0 there is an $n_{0}$ for which for all $n>n_{0}$ is...

$\displaystyle a_{n} < \frac{h}{n}$
This means
$\forall h>0\,\exists n_0\,\forall n>n_0\;a_n<\frac{h}{n}\qquad(1)$

If they exist an $h > 0$ and an $n_{0}$ such that for any $n>n_{0}$ is $a_{n} \ge \frac{h}{n}$, then the series diverges
And this means
$\exists h>0\,\exists n_0\,\forall n>n_0\;a_n\ge\frac{h}{n}\qquad(2)$
I agree that (2) implies that the series diverges, i.e., a contradiction. (Even then, how does it imply that? Using the fact that the harmonic series diverges? And how to prove that? Using the integral test? The original claim can be proved using just the fact that partial sums of a convergent series form a Cauchy sequence, without using such complicated concepts as Riemann integral.) But (2) is not the negation of (1). It is stronger than the negation of (1), so it is easier to derive a contradiction from (2), but it does not mean that the negation of (1) is false.