Why this expression is true

1. May 3, 2009

transgalactic

$$\sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}$$

ln is not always bigger then 1
so when i am doing the comparing test
i cant use that
because ln 1 =0

??

Last edited: May 3, 2009
2. May 3, 2009

Staff: Mentor

ln n > 1 for almost every positive number n. Compare the graphs of y = ln x and y = 1 and you'll see that what I'm saying is true.

3. May 4, 2009

transgalactic

you said yourself "almost" not absolutely

4. May 4, 2009

diazona

Remember that that sum is really just a whole list of terms, all added up. So you can split the sum into two sums, or write out some of the terms explicitly if you want. Just to give an example:

$$\sum_{n=0}^{\infty} n^2 e^{-n} = \left(\sum_{n=0}^{6} n^2 e^{-n}\right) + \left(\sum_{n=7}^{\infty} n^2 e^{-n}\right) = 0^2 e^{-0} + 1^2 e^{-1} + 2^2 e^{-2} + \sum_{n=3}^{\infty} n^2 e^{-n}$$

You can do something like this for one or both of the sums in your expression, and it should help.

5. May 4, 2009

Staff: Mentor

In an infinite series, you can always ignore a finite number of terms without affecting whether the series converges or diverges.