Is this the formal way of comparing series?

[flyingpig]

Homework Statement

Determine whether \sum_{n=2}^{\infty}\frac{1}{n^2 \ln(n)} converge or diverge.

The Attempt at a Solution

If you are a marker, what is the formal work you want to see? This is how I would write it on my paper

Imagine below is a paper

\sum_{n=2}^{\infty}\frac{1}{n^2 \ln(n)} < \sum_{n=2}^{\infty}\frac{1}{n^2 }

\sum_{n=2}^{\infty}\frac{1}{n^2 } is a p-series with p > 1, so it converges and by the comparison test, \sum_{n=2}^{\infty}\frac{1}{n^2 \ln(n)} must also converge

 

[SammyS]

I think you meant that your p-series converges.

 

[flyingpig]

I think you meant that your p-series converges.

Yes...*goes fixing*

Why do I keep making these mistakes lol, it's like my physics final two days ago...

 

[LCKurtz]

You have the general idea. But, for full credit, you need to note that your presumed termwise comparison

<br /> \frac{1}{n^2 \ln(n)} &lt; \frac{1}{n^2 }<br />

is not true when n = 2 and observe that that doesn't matter.

 

[SammyS]

I would rather see \frac{1}{n^2 \ln(n)} &lt; \frac{1}{n^2 }\,,\text{ for } n &gt; 1\ .<br />

 

[flyingpig]

So don't write the summation, just write out the term, a_n and b_n

 

[SammyS]

Yes, compare terms.

 

[flyingpig]

Actually wait, LCKurtz made a good point, just because one term is bigger, doesn't mean the sum is. So it doesn't matter in the end right?

 

[SammyS]

But if each term of one series is larger than each corresponding term in the other series, and they're all positive ...

 

[LCKurtz]

But if each term of one series is larger than each corresponding term in the other series, and they're all positive ...

But it isn't, not for n = 2. Easily taken care of but it requires mentioning.

 

[SammyS]

Point well taken.