# Comparison test

1. Feb 28, 2015

### yuk

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

determine series convergence of divergence

summation (n=1 to infinity) n/n^2 +1

2. Relevant equations

3. The attempt at a solution
I take the limit comparison
limit (1/n)/ (n/(n^2 +1) =1
for 1/n if i use p series the series diverge
if i use the method to take limit of sequence An then 1/n =0 so the series converge

2. Feb 28, 2015

### LCKurtz

$lim_{n\to\infty}a_n=0$ does not imply convergence.

3. Feb 28, 2015

### yuk

but In the text book
nTh term test
if the sequence An converge to zero, then the series An converges

I was so confusing

4. Feb 28, 2015

### SammyS

Staff Emeritus
You should have the denominator in parentheses.

n/(n^2 +1)

You may be referring to the ratio test:

Suppose $\ \displaystyle \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = r \, .$
Then:
if r > 1, the series diverges
if r < 1, the series converges
if r = 1, the test is inconclusive. (appears to be the case here.)​

Use the limit comparison test.

5. Mar 1, 2015

### Staff: Mentor

That's not what it says.
The nth term test for divergence says something along these lines.
If a series $\sum a_n$ converges, then $\lim_{n \to 0} = 0$

The converse of this statement (i.e., if $\lim_{n \to 0} = 0$, then the series $\sum a_n$ converges) IS NOT TRUE!! A classic example is the series $\sum 1/n$. Even though $\lim_{n \to \infty} 1/n = 0$, the series itself diverges.

An equivalent way to state the nth term test for divergence is this:
If $\lim_{n \to 0} a_n \neq 0$, then the series $\sum a_n$ diverges.