integral test

**tsal15** · January 10th 2009, 04:45 AM

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

**flyingsquirrel** · January 10th 2009, 04:56 AM

Hi,

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

)

**mr fantastic** · January 10th 2009, 04:59 AM

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 .....

**Prove It** · January 10th 2009, 05:01 AM

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?

**mr fantastic** · January 10th 2009, 05:08 AM

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.

**Prove It** · January 10th 2009, 05:13 AM

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...

**Krizalid** · January 10th 2009, 04:53 PM

Always verify that $f$ must be positive, continuous and decreasing on $[1,\infty[.$

**tsal15** · January 10th 2009, 05:10 PM

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

**tsal15** · January 13th 2009, 05:54 AM

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?

**Chop Suey** · January 13th 2009, 06:34 AM

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.

**mr fantastic** · January 13th 2009, 11:12 AM

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.

**Prove It** · January 13th 2009, 02:56 PM

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?

Thread: integral test

LinkBack

Thread Tools

Display

integral test

Similar Math Help Forum Discussions

Integral Test and comparison test help!

Integral Test

Integral Test

Integral Test/Comparison Test

Integral Test

Search Tags