# Thread: why this expression is true..

1. ## why this expression is true..

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

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

??

2. Originally Posted by transgalactic
$\displaystyle \sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}$

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

??
because $\displaystyle \ln{n} > 1$ for $\displaystyle n \geq 3$

3. Originally Posted by transgalactic
$\displaystyle \sum_{n=1}^{\infty}\frac{\ln n}{n}> \sum_{n=1}^{\infty}\frac{1}{n}$

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

??
But $\displaystyle \sum_{n=3}^{\infty}\frac{\ln n}{n}> \sum_{n=3}^{\infty}\frac{1}{n}$

is true so you can use the comparision test on that series and add the first couple of terms. Adding a few terms to the series won't change its convergenece (divergence).

4. Nevermind that.

ln(n)/n may initally be smaller, but when n > e it overtakes 1/n.

Seeing as both terms are over some number n, the determining factor is the size of the numerator. 1 will always be 1, but ln(n) is always increasing.

5. More accurately, ln n is bigger than 1 when n is bigger than e (or 2.718)