# Nth term test for divergence

## Homework Statement

[/B]
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

## Homework Equations

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

## 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

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

Thank you,

SY
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.

• SYoungblood
HallsofIvy
Homework Helper
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.

• SYoungblood
Mark44
Mentor

## Homework Statement

[/B]
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

## Homework Equations

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

## 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
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).

• 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

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

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

• SYoungblood
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".