Solve Euler's Constant Homework: Laws of Logs & Calculus Facts

[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, 1/(n+1)&lt; 1/x[/itex] so <br /> \int_n^{n+1} 1/(n+1)dx= (1/(n+1))\int_n^{n+1}dx= 1/(n+1)&amp;lt; \int_n^{n+1} 1/x dx.

 

[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)