1. ## integral test

I just want to quickly check....

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

2. Hi,
Originally Posted by tsal15
I used the integral test to check if $\displaystyle \sum_{n=2}^{\infty}\frac{1}{nln(n)}$converges.... and i found that it does.
No it does not, this series is divergent. (it can be shown using the integral test )

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

I used the integral test to check if $\displaystyle \sum_{n=2}^{\infty}\frac{1}{nln(n)}$converges.... and i found that it does. Can someone verify this? Thanks
$\displaystyle \sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx$ and I'm pretty confident that the integral does not have a finite value .....

4. Originally Posted by mr fantastic
$\displaystyle \sum_{n=2}^{\infty}\frac{1}{n \ln(n)} > \int_2^{+\infty} \frac{1}{x \ln x} \, dx$ and I'm pretty confident that the integral does not have a finite value .....
Are you sure that sign is around the right way?

5. 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.

6. 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...

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

8. Originally Posted by Krizalid
Always verify that $\displaystyle f$ must be positive, continuous and decreasing on $\displaystyle [1,\infty[.$
Yes when using the integral test

Thanks for the reminder

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

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

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

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

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

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

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

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

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

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

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

ln(ln(2)) is a value. Thus even if $\displaystyle ln(ln(\infty))$ 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.

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

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

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

ln(ln(2)) is a value. Thus even if $\displaystyle ln(ln(\infty))$ does diverge, ln(ln(2)) will converge..... so does this change things?
In the grand scheme of things, $\displaystyle \ln{\ln{2}}$ 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?