# Divergence of 1/k

1. Apr 16, 2015

### RyanTAsher

1. The problem statement, all variables and given/known data

If lim(k>inf) 1/k, goes to 0, why does it diverge?

2. Relevant equations

Divergent series test

3. The attempt at a solution

i dont understand why 1/k (harmonic series) diverges, when according to the divergent series test, it should converge to 0.

2. Apr 16, 2015

### WWGD

What is the "divergence series test"? Maybe I(we) know it by a different name? But the n-th term going of the series going to 0 is definitely necessary, but not sufficient for convergence of the series.

3. Apr 16, 2015

### RyanTAsher

Yes that is the same test. So is it just a confirmation for divergence, but inconclusive at L = 0?

4. Apr 16, 2015

### LCKurtz

"It" doesn't diverge. $\frac 1 k \to 0$.

Be careful about distinguishing $\frac 1 k$ with the series $\sum \frac 1 k$. The series and the sequence of its terms are different things.

5. Apr 16, 2015

### RyanTAsher

I'm sorry, I should've paid attention to the wording of my question. It should have been, why doesn't the series $\sum \frac 1 k$ converge? I was under the impression that if the limit of the series goes to 0, then the series converges. Yet if it doesn't it diverges. If $\lim_{k\to\infty} \sum\limits_{k=1}^{\infty} \frac 1 k$ = 0, then it converges? I don't really understand.

6. Apr 17, 2015

### LCKurtz

That doesn't make any sense. A series either converges, in which case it has a sum, or it diverges. There is no "limit of a series".

Again, that makes no sense. The $k$ in the summation is a dummy variable. There is no $k$ to take the limit of.

You are still confusing the limit of the $k$th term with something else. The relevant theorem is : If the series $\sum a_k$ converges, then $a_k\to 0$. The contrapositive statement is that if $a_k$ doesn't converge to $0$, the series diverges, which is why it is sometimes called a divergence test.

What the theorem doesn't say is that if $a_k\to 0$ then $\sum a_k$ converges. That is the converse of the theorem and it isn't true.

7. Apr 17, 2015

### Staff: Mentor

Your limit makes no sense, since in the sum, k takes on values 1, 2, 3, ... and so on.

Take another look at the wording of the theorem you are using -- the mistake you are making is one that many students make. For a series $\sum_{k = 1}^{\infty} a_k$, the Nth Term Test for Divergence, as it is usually called, says that if lim ak ≠ 0, the series diverges. The mistake that many new students make is in thinking that if if lim ak = 0, the series converges. THIS IS NOT TRUE, and has no connection with the Nth Term Test for Divergence. All it does is say whether a series diverges or not. It DOES NOT tell you that a given series converges.

Edit: LCKurtz beat me to it. The form of this theorem he cited is equivalent to what I wrote.

8. Apr 17, 2015

### Ray Vickson

You have seriously misunderstood an important fact about infinite series: for convergence of $\sum_{n=1}^{\infty} a_n$ it is necessary to have $a_n \to 0$ as $n \to \infty$. However, that is not sufficient; in fact, the series $\sum 1/n$ is a good counterexample.

There are lots of proofs of divergence, but the easiest one is based on comparison with an integral. We have $1/1 \geq 1/x$ for $1 \leq x \leq 2$, $1/2 \geq 1/x$ for $2 \leq x \leq 3$, $\ldots$. In general, $1/k \geq 1/x$ for $k \leq x \leq k+1$. Thus, $1/1 \geq \int_1^2 dx/x$, $1/2 \geq \int_2^3 dx/x$, etc, so
$$\sum_{n=1}^N \frac{1}{n} \geq \left( \int_1^2 + \int_2^3 + \cdots + \int_N^{N+1} \right) \frac{dx}{x} = \int_1^{N+1} \frac{dx}{x} = \ln(N+1)$$
This implies $\sum_{n=1}^N 1/n \to \infty$ as $N \to \infty$, so the series diverges.

9. Apr 17, 2015

### RyanTAsher

Thank you, all of you. I finally understand. I guess that's the question I was trying to state all along when I said

I just meant, "Does it prove the series diverges, but not necessarily that the series converges if L = 0"

Thank you again for your time.