Divergence of Series Summation (n=1 to infinity) n/n^2 +1

[yuk]

Homework Statement

determine series convergence of divergence

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

Homework Equations

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

The answer is diverge

 

[LCKurtz]

Homework Statement

determine series convergence of divergence

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

Homework Equations

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

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

 

[yuk]

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

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

I was so confusing

 

[SammyS]

Homework Statement

determine series convergence of divergence

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

You should have the denominator in parentheses.

n/(n^2 +1)

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

I was so confusing

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.

 

[Mark44]

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

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.