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.