Series/logarithm questions

Hi, I was wondering how to solve these 2 questions:

c) use the proof from part a) and d) [e^x=>x+1 and (1+1)(1+1/2)...(1+1/n)=n+1] to prove that e^(1+1/2+1/3+...+1/n)>n

d) Find a value of n for which 1+1/2+1/3+...+1/n>100

You don't have to provide the entire answer, but some hints or part of the work would be nice.

Thanks in advance!

Re: Series/logarithm questions

Having established that:

$\displaystyle (1+1)\left(1+\frac{1}{2} \right)\left(1+\frac{1}{2} \right)\cdots\left(1+\frac{1}{n} \right)=n+1$

and:

$\displaystyle e^x\ge x+1$

we may write:

$\displaystyle e^{1}\ge 1+1$

$\displaystyle e^{\frac{1}{2}}\ge 1+\frac{1}{2}$

$\displaystyle e^{\frac{1}{3}}\ge 1+\frac{1}{3}$

$\displaystyle \vdots$

$\displaystyle e^{\frac{1}{n}}\ge 1+\frac{1}{n}$

Since all values are positive, we may multiply down. What do you get?

d) Having established that:

$\displaystyle e^{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}> n$

this implies:

$\displaystyle 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}>\ln(n )$

Now, let:

$\displaystyle \ln(n)=100$

What does this say about $\displaystyle n$?

Re: Series/logarithm questions

Oh...thanks so much. I can't believe I didn't think that way for c) :( disappointed in myself lol