Euler's Constant

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.


Science Advisor
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
[tex]\int_n^{n+1} 1/(n+1)dx= (1/(n+1))\int_n^{n+1}dx[/tex][tex]= 1/(n+1)< \int_n^{n+1} 1/x dx[/tex].
I can't tell if I am reading this wrong, but for part (b), [itex]T_{2n}-T_n[/itex] i get [itex]\sum^{2n}_{r=1} \frac{1}{r}-\sum^{n}_{r=1}\frac{1}{r}-log2[/itex]. I was thinking i had the summations wrong because i am definitely not ending up with [itex]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...=log2[/itex]
Last edited:
thanks Ivy
Can someone tell me if [itex]\sum^{2n}_{r=n} = \sum^{2n}_{r=1}-\sum^{n}_{r=1}[/itex] ? This is what I am trying to work with. I was told I need to rearrange, but I dont think I am rearranging the correct sequence of numbers. I dont see how [itex]\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}[/itex] can turn into [itex]\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....[/itex]

PS: this is for part (b)

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