Proving Inequalities of Euler-Mascheron Constant with Taylor Expansion

[kde2520]

Homework Statement

With n>1, show that (a) \frac{1}{n}-ln\frac{n}{n-1}<0
and (b) \frac{1}{n}-ln\frac{n+1}{n}>0

Use these inequalities to show that the Euler-Mascheron constant (eq. 5.28 - page330) is finite.

Homework Equations

This is in the chapter on infinite series, in the section on Taylor Expansion, so I guess Taylor, Maclaurin, and Binomial theorem are fair game.

The Attempt at a Solution

I first wrote the logarithm as a difference of logs and then tried to expand them in the Maclaurin series. But that apparently doesn't work since ln(0) and 1/0 are undefined...

I also don't understand the statement at the end. Is that supposed to be a hint or a third part to the problem?

Any help would be great, thanks.

 

[lurflurf]

For the first two part change the log to log(1+x) so that we want to show
x/(1+x)<log(1+x)<x
for |x|<1
this can be done by using an integral or series representation for log
The third bit is indeed a part of the problem and not a hint, infact one could say the first two parts are a hint for the third
The constant in question is
lim_{n->infinity} [1/1+1/2+1/3+...+1/n-log(n)]
use the given inequalities to bound the constant

 

[kde2520]

I'm not exactly sure how to go about changing the log to log(1+x). Do you mean to factor an x out of the demoninator for the first one and then have log(1-1/x), and for the second one log(1+1/x)?

 

[HallsofIvy]

I think he means let x= n/(n-1) so that ln(n/(n-1)) becomes just ln x. Of course, then x(n-1)= xn- x= n so n(x-1)= x and n= x/x-1 so that 1/n= (x-1)/x= 1- 1/x. Then your inequality 1/n- ln(n/(n-1))<0 becomes 1- 1/x- ln(x)< 0.

 

[kde2520]

Ok, but I'm not seeing how the two approaches are connected since lurflurf said to change the logarithm to ln(1+x).

 

[lurflurf]

find that log(1+x)~x (x~0)is a simple form to work with
log(x)~-1+x (x~1) amounts to the same thing, matter of taste

we have
1/n-log(n/(n-1))
and
1/n-log((n+1)/n)

n/(n-1)=1+1/(n-1)
let
x=1/(n-1)
n/(n-1)=1+1/(n-1)->1+x
1/n=1/(n-1+1)=[1/(n-1)]/[1+1/(n-1)]->x/(1+x)
1/n-log(n/(n-1))->x/(1+x)-log(1+x)

(n+1)/n=1+1/n
let x=1/n
(n+1)/n=1+1/n->1+x
1/n-log((n+1)/n)->x-log(1+x)

we desire to show
x-log(1+x)>0
and
x/(1+x)-log(1+x)<0
or
x/(1+x)<log(1+x)<x
|x|<1
This is easily done by any number of methods including using series
log(1+x)=x-x/2+x^3/3-x^4/4+...
or integrals
log(1+x)=int(1/(1+t),t,0,x)

once the hint inequalitites are verified use them to bound the constant

 

[kde2520]

Ok, that helps a lot. Thanks!