# Show that a Series Diverges

Question:

Let a sequence ($a_n$) have the property $\lim \limits_{n \to \infty} na_n = a > 0$

Show that the series $\sum_{n=1}^\infty a_n$ diverges

Attempts:

Basically, I firstly tried separating the left-hand side so,

$(\lim \limits_{n \to \infty} n)( \lim \limits_{n \to \infty} a_n) = a$

$\implies\lim \limits_{n \to \infty} a_n = \frac{a}{\lim \limits_{n \to \infty} n} = 0$

But this implies that the series $\sum_{n=1}^\infty a_n$ has satisfied the vanishing test thus at this point we cannot determine whether it is convergent or divergent.

As for my other informal attempt is that since $\lim \limits_{n \to \infty} na_n = a$, that implies the rate of growth of the sequence ($a_n$) must be identical to the rate of growth of the sequence $b_n = n$. Therefore, the general term of $a_n$ must be in the form of $k\frac{1}{n}$ where $k \in \Bbb R$. Finally, since $\sum_{n=1}^{\infty} \frac{k}{n}$ is the harmonic series, therefore the series diverges. (The harmonic series also satisfies my first attempt equality).

At least to me, my first attempt looks fine but it leads me to nowhere while my second attempt is not rigorous enough. So basically what I am asking is my second attempt sufficient/rigorous enough (because it doesn't seem so) and is there another more easier method to prove that the series diverges?

EDIT

New method:

$\frac{\lim \limits_{n \to \infty} na_n}{n} > \frac{0}{n}$

$\lim \limits_{n \to \infty} \frac{na_n}{n} > 0$

$\lim \limits_{n \to \infty} a_n > 0$

Thus the series $\sum_{n=1}^{\infty} a_n$ fails the vanishing test so it is divergent.

It 'works' but I feel like there is a fundamental flaw in that proof somewhere (moving from line 1 to line 2). Especially since this proof is incorrect in the case $a_n = \frac{k}{n}$

• Re: edit. No. You're thinking of $n$ as fixed downstairs in the first line, but not in the second. – David Mitra Nov 26 '14 at 9:04
• You might be interested by this result: if $a_n$ and $b_n$ are two real positive sequences such as : $a_n$ ~ $b_n$, then: $\sum a_n$ is of the same nature (that is, converge or diverge) as $\sum b_n$. – mvggz Nov 26 '14 at 10:44

If you were an experienced mathematician, your second attempt would be perfect. If you are not, you should motivate more carefully your conclusion. For instance, you could say that, eventually, $$a_n \geq \frac{a}{2n},$$ and therefore $$\sum_n a_n \geq C +\sum_n \frac{a}{2n},$$ which diverges.

• I assume you mean "you could say that, eventually, ..." rather than "definitely"? – Jason Nov 26 '14 at 9:02
• The lower bound by $C$ times a sum is not what one deduces naturally from the steps before. – Did Nov 26 '14 at 10:06
• @Jason Sorry, I'm italian and we say "definitivamente" ;-) – Siminore Nov 26 '14 at 12:05
• Apologies for the couple of months delay on choosing the best answer. Also, my gratitude to you and Barry Cipra for answering my questions =). I was very impressed with the speed you answered my question (literally in couple of minutes after I posted it) =). – OthmanEmpire Feb 3 '15 at 21:26

This answer addresses only the "new method" in the OP's edit. That approach is fundamentally flawed in line 1, and consequently in the (implicit) logic that moves from line 1 to line 2 to line 3.

The problem with line 1 is that you're using the symbol $n$ in two separate ways: In the numerator on the left hand side it's a variable to define a limit, while in the denominator on both sides it's a variable representing an arbitrary but fixed value. A proper way to write that inequality would be

$${\lim_{k\to\infty}ka_k\over n}\gt{0\over n}$$

The twin use of the symbol $n$ is what disrupts the logic of moving from line 1 to line 2. It's perfectly OK to say

$${\lim_{k\to\infty}ka_k\over n}=\lim_{k\to\infty}\left({ka_k\over n}\right)$$

since the variable $n$ is a constant as far as the limit is concerned. But writing things properly, with separate symbols, makes it clear there's no cancellation going on here, so you can't get to line 3.

The moral of the story is, be careful about your use of symbols. Once you've chosen to use a symbol for one purpose, it's no longer available for other use in the same expression.