Proving an inequality for Euler's constant with logarithms

[mynameisfunk]

Homework Statement

For this problem, we will use the basic laws of logarithms and calculus facts about the natural logarithm $log(x)$, even though we haven't proven them in our class yet.

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

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

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

Homework Equations

The Attempt at a Solution

OK, I have only started the first part of a., I am still just trying to show that $\frac{1}{n+1} \leq \int_n^{n+1} \frac{1}{x} dx$. Already running into trouble... Here is what I am trying but for some reason this won't work out:
$$\frac{1}{n+1} \leq log(\frac{n+1}{n})$$
$$\frac{1}{n+1} \leq log(1+\frac{1}{n})$$
$$ee^{\frac{1}{n}} \leq 1+ \frac{1}{n}$$
Ive gone wrong but I can't see it. I know this won't hold.

 

[HallsofIvy]

You are going the wrong way. You don't want to do the integration to find a bound, you want to find a bound so you don't have to do the integration!

For all x between n and n+1, [tex]1/(n+1)< 1/x[/itex] so <br /> $$\int_n^{n+1} 1/(n+1)dx= (1/(n+1))\int_n^{n+1}dx$$$$= 1/(n+1)< \int_n^{n+1} 1/x dx$$.[/tex]

 

[mynameisfunk]

I can't tell if I am reading this wrong, but for part (b), $T_{2n}-T_n$ i get $\sum^{2n}_{r=1} \frac{1}{r}-\sum^{n}_{r=1}\frac{1}{r}-log2$. I was thinking i had the summations wrong because i am definitely not ending up with $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...=log2$

 

[mynameisfunk]

thanks Ivy

 

[mynameisfunk]

Can someone tell me if $\sum^{2n}_{r=n} = \sum^{2n}_{r=1}-\sum^{n}_{r=1}$ ? This is what I am trying to work with. I was told I need to rearrange, but I don't think I am rearranging the correct sequence of numbers. I don't see how $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ can turn into $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$


PS: this is for part (b)