# integral test

• January 10th 2009, 04:45 AM
tsal15
integral test
I just want to quickly check....

I used the integral test to check if $\sum_{n=2}^{\infty}\frac{1}{nln(n)}$converges.... and i found that it does. Can someone verify this? Thanks :)
• January 10th 2009, 04:56 AM
flyingsquirrel
Hi,
Quote:

Originally Posted by tsal15
I used the integral test to check if $\sum_{n=2}^{\infty}\frac{1}{nln(n)}$converges.... and i found that it does.

No it does not, this series is divergent. (it can be shown using the integral test :D)
• January 10th 2009, 04:59 AM
mr fantastic
Quote:

Originally Posted by tsal15
I just want to quickly check....

I used the integral test to check if $\sum_{n=2}^{\infty}\frac{1}{nln(n)}$converges.... and i found that it does. Can someone verify this? Thanks :)

$\sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx$ and I'm pretty confident that the integral does not have a finite value .....
• January 10th 2009, 05:01 AM
Prove It
Quote:

Originally Posted by mr fantastic
$\sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx$ and I'm pretty confident that the integral does not have a finite value .....

Are you sure that sign is around the right way?
• January 10th 2009, 05:08 AM
mr fantastic
Quote:

Originally Posted by Prove It
Are you sure that sign is around the right way?

All the left hand rectangles of width 1 (starting from x = 2) have areas lying above the corresponding area under the curve so I'm pretty sure I'm sure.

But since I'm about to go to bed I will leave it to others to point out my mistakes if there are any.
• January 10th 2009, 05:13 AM
Prove It
Quote:

Originally Posted by mr fantastic
All the left hand rectangles of width 1 (starting from x = 2) have areas lying above the corresponding area under the curve so I'm pretty sure I'm sure.

But since I'm about to go to bed I will leave it to others to point out my mistakes if there are any.

Yes that would make sense.

Plus, to show that it's divergent it WOULD have to be greater than something tending to infinity wouldn't it...
• January 10th 2009, 04:53 PM
Krizalid
Always verify that $f$ must be positive, continuous and decreasing on $[1,\infty[.$ :)
• January 10th 2009, 05:10 PM
tsal15
Quote:

Originally Posted by Krizalid
Always verify that $f$ must be positive, continuous and decreasing on $[1,\infty[.$ :)

Yes when using the integral test :)

Thanks for the reminder :)
• January 13th 2009, 05:54 AM
tsal15
Quote:

Originally Posted by mr fantastic
$\sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx$ and I'm pretty confident that the integral does not have a finite value .....

Ok now, I've got a problem...

the integral equals => $ln(ln(x)$ with infinity and 2 being the bounds

therefore, $ln(ln(\infty)) - ln(ln(2))$

ln(ln(2)) is a value. Thus even if $ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?
• January 13th 2009, 06:34 AM
Chop Suey
Quote:

Originally Posted by tsal15
Ok now, I've got a problem...

the integral equals => $ln(ln(x)$ with infinity and 2 being the bounds

therefore, $ln(ln(\infty)) - ln(ln(2))$

ln(ln(2)) is a value. Thus even if $ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?

Nope, the integral diverges.
• January 13th 2009, 11:12 AM
mr fantastic
Quote:

Originally Posted by tsal15
Ok now, I've got a problem...

the integral equals => $ln(ln(x)$ with infinity and 2 being the bounds

therefore, $ln(ln(\infty)) - ln(ln(2))$

ln(ln(2)) is a value. Thus even if $ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?

You're interested in the final value of the integral, not the individual bits that contribute to calculating this value.
• January 13th 2009, 02:56 PM
Prove It
Quote:

Originally Posted by tsal15
Ok now, I've got a problem...

the integral equals => $ln(ln(x)$ with infinity and 2 being the bounds

therefore, $ln(ln(\infty)) - ln(ln(2))$

ln(ln(2)) is a value. Thus even if $ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?

In the grand scheme of things, $\ln{\ln{2}}$ is very small when compared to infinity.

So even though this value is being subtracted, it's still close enough to infinity to say that the integral tends to infinity.

Therefore it diverges.

Make sense?