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**1. The problem statement, all variables and given/known data**

For this problem, we will use the basic laws of logarithms and calculus facts about the natural logarithm [itex]log(x)[/itex], even though we havent proven them in our class yet.

(a) Explain why [itex]\frac{1}{1+n} \leq \int_n^{n+1} \frac{1}{x}dx[/itex]. Then, setting [itex]T_n= \sum^n_{r=1} \frac{1}{r}-logn[/itex], show that [itex]0\leqT_{n+1}\leqT_n\leq1[/itex], for all [itex]n[/itex]. Conclude that [itex]\gamma = \lim{x\rightarrow0}T_n[/itex] exists. This constant is known as Euler's Constant. It is not even known whether \gamme is rational or not.

(b) Consider [itex]T_{2n}-T_n[/itex] and show that [itex]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...=log2[/itex]

(c) Consider [itex]T_{4n}-\frac{1}{2}T_{2n}-\frac{1}{2}T_n[/itex] and show that [itex]1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...=\frac{3}{2}log2[/itex]

**2. Relevant equations**

**3. The attempt at a solution**

OK, I have only started the first part of a., I am still just trying to show that [itex]\frac{1}{n+1} \leq \int_n^{n+1} \frac{1}{x} dx[/itex]. Already running into trouble... Here is what I am trying but for some reason this wont work out:

[tex]\frac{1}{n+1} \leq log(\frac{n+1}{n})[/tex]

[tex]\frac{1}{n+1} \leq log(1+\frac{1}{n})[/tex]

[tex]ee^{\frac{1}{n}} \leq 1+ \frac{1}{n}[/tex]

Ive gone wrong but I can't see it. I know this wont hold.