Prove that $\lim na_n=0$

Question

Suppose that the positive series $\sum\limits_{n=1}^{\infty} a_n$ converges and $a_n>a_{n+1}$ for all $ n\in N$. Prove that $\lim na_n=0$

This is my proof. Suppose that $\lim na_n=a$, with $a>0$. Then we have $\lim \dfrac{a_n}{\frac{a}{n}}=1$. Since $\sum\limits_{n=1}^{\infty} \frac{a}{n}$ diverges then $\sum\limits_{n=1}^{\infty} a_n$ must diverge. We have contradition, thus $\lim na_n=0$ (since $\sum\limits_{n=1}^{\infty} a_n$ is a positive series)

In my proof, I didn't use the hypothesis $a_n>a_{n+1}$. I wonder if my proof is wrong or that hypothesis is useless.

Answer 1

Your proof is incomplete because the limit $\lim\limits_{n\to \infty} na_n$ may not exist. You would not, in your proof, be able to apply the limit comparison test in the case that $\lim\limits_{n\to \infty} na_n$ does not exist.

To prove the result, we show that the even and odd parts of $na_n$ have limit zero. Let $s_n$ be the sequence of partial sums of $\sum a_n$, and consider the even part, $2na_{2n}$:

$$2na_{2n} = 2(\underbrace{{a_{2n}+\cdots + a_{2n}}}_{\text{$n$ times}}) < 2(a_{n+1}+ a_{n+2} + \cdots + a_{2n}) = 2(s_{2n} - s_n)$$

Since $0 < 2na_{2n} < 2(s_{2n} - s_n)$ for all $n$ and $\lim\limits_{n\to \infty} (s_{2n} - s_n) = 0$, then by the squeeze theorem $\lim\limits_{n\to \infty} 2na_{2n} = 0$. Now consider the odd part, $(2n+1)a_{2n+1}$. Since $a_{2n+1} < a_{2n}$, then $(2n + 1)a_{2n+1} < 2na_{2n} + a_{2n+1}$. Since $\lim\limits_{n\to \infty} 2na_{2n} = 0$ and $\lim\limits_{n\to \infty} a_{2n+1} = 0$ (since $\sum a_n$ converges), then $\lim\limits_{n\to \infty} (2na_{2n} + a_{2n+1}) = 0$. Again, by the squeeze theorem, $\lim\limits_{n\to \infty} (2n+1)a_{2n+1} = 0$.

Answer 2

The reasoning of your proof follows: "if the limit $\lim_{n \to \infty}na_n$ exists and is equal to some $a$, then $a = 0$. This tells us nothing of the truth of the statement because you need to prove that the first part is true i.e. the limit actually exists.

To prove that $\lim_{n \to \infty}na_n$ exists, we will have to consider what we have available to us:

1. $\sum_{n=1}^{\infty} a_n$ converges.
2. For any $n$, we have $a_{n} > a_{n+1}$.

Since the partial sums converge by list item 1, by the Cauchy criterion, for any $\epsilon\over 2$ from the devil, there exists an $N_0$ such that for any $n > N_0$ and $p \geq 1$, we have that:

$$ \left \lvert\sum_{i=n}^{n+p}a_i \right \rvert < \frac{\epsilon}{2} $$

in other words, there is some point $N_0$ after which you have:

$$ \left \lvert a_n + a_{n+1} + \cdots + a_{n+p} \right \rvert < \frac{\epsilon}{2} $$

for any choice of $n > N_0$ and $p \geq 1$.

Now we want to introduce the $na_n$ sequence into this inequality somehow and maybe bound it. What can you say about the summation? Well the sequence is decreasing so we have $a_{n+p} < a_{n+i}$ for all $i < p$. So with our inequality, we get:

$$ \left \lvert pa_{n+p} \right \rvert = \left \lvert a_{n+p} + a_{n+p} + \cdots + a_{n+p} \right \rvert < \left \lvert a_n + a_{n+1} + \cdots + a_{n+p} \right \rvert < \frac{\epsilon}{2} $$

Since we want to investigate the sequence $na_n$ we can set $p = n$ and multiply everything by $2$ to get:

$$ \left \lvert 2na_{2n} \right \rvert < 2 \left \lvert a_n + a_{n+1} + \cdots + a_{2n} \right \rvert < \epsilon \\ $$

So we have bounded the sequence $2na_{2n}$ by $\epsilon$ and thus it converges to $0$. However $2na_{2n}$ is only a subsequence i.e. just the even terms of $na_n$ so we aren't done. It is known that if both the even and odd terms of a sequence converge to the same limit, then the entire sequence converges to that limit.

We have proved the even terms converge to $0$. Can you extend this to the odd terms? What can you say about the sequence $(2n+1)a_{2n+1}$ in terms of what it must be smaller or bigger than? Hint: use the information from the bullet points.