Why Does the Series Sum of (ln n)/n Exceed Sum of 1/n?

[transgalactic]

<br /> \sum_{n=1}^{\infty}\frac{\ln n}{n}&gt; \sum_{n=1}^{\infty}\frac{1}{n}<br />

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

??

 

[Mark44]

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.

 

[transgalactic]

you said yourself "almost" not absolutely

 

[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.

 

[Mark44]

you said yourself "almost" not absolutely

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