# Nth term test for divergence

1. Jun 4, 2015

### SYoungblood

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

Here is an nth term test for determining divergence, I think I have it, but wanted another opinion --

1/34 + 1/35 + 1/36+ … + 1/1,000,034 -- I

2. Relevant equations

∑(upper limit ∞)(lower limit n=0) 1/(n+34)

3. The attempt at a solution

1/34 + 1/35 + 1/36+ … + 1/1,000,034 -- I see this eventually hits zero when the numbers get very large.

for a series ∑n=1∞an, if lim n→∞ of an ≠ 0, then the series diverges.

So, the nth term test fails because we have a convergent limit at zero, am I right?

Thank you,

SY

2. Jun 4, 2015

### David Carroll

Exactly. A zero limit is necessary but not sufficient to determine convergence.

Also: this series diverges because it is simply the harmonic series minus the first 33 terms.

3. Jun 4, 2015

### HallsofIvy

Staff Emeritus
Be sure that you understand that the "divergence test" says that $$\sum_0^\infty a_n$$ diverges if $$a_n$$ does not go to 0. If the sequence does go to 0, as is the case here, there is no conclusion. You cannot draw any conclusion as to divergence.

4. Jun 4, 2015

### Staff: Mentor

You are misunderstanding the Nth Term Test, which is a very common mistake for calculus students.

In one form, the theorem says this:
"In a series $\sum a_n$, if the series converges, then $\lim_{a_n} = 0$."
An equivalent form is the converse, which is "In a series $\sum a_n$, if $\lim_{a_n} \neq 0$, then the series diverges.
The theorem says absolutely nothing about a series for which $\lim_{a_n} = 0$. There are loads of series for which the nth term approaches zero. Some of the converge (e.g., geometric series) and some of the diverge (e.g., harmonic series).

5. Jun 4, 2015

### SYoungblood

The funny thing was, I told myself I wouldn't make that mistake several times.

So, of course I did. Calculus is a humbling adversary.

I wanted to try one more -- ∑(upper limit ∞)(lower limit n=0) cos (9/n)

I have the nth term test failing because the limit approaches zero as n gets huge, and the cosine equalling one.

I got this problem out of a well-known text and am just trying to make myself a little better.

SY

6. Jun 4, 2015

### pasmith

I suspect you mean the lower limit to be $n =1$, since $\cos(\frac 90)$ is not defined.

Correct.

7. Jun 6, 2015

### David Carroll

I think that's what SYoungblood meant: the test doesn't apply because an has to approach zero in order to converge to infinity but that it's not enough (i.e. another test such as the "divergent series minus a finite number of terms"-test), so the test "fails".