Friday, January 1, 2016

calculus - Bounding ln(1+1/x) for x>0




I am hoping to somewhat rigorously establish the bound
1x+1/2<ln(1+1/x)<1/x
for any x>0. The upper bound is clear since
$$
1+1/x$$
The lower bound seems dicier. A taylor expansion argument wouldn't seem to work near zero, since 1/x blows up.




Geometrically at least it seems clear near zero. Since ln(1+1/x) is decreasing and convex, and limx0+ln(1+1/x)=+ and at 0
11/2=2
I can establish the lower bound near the origin. Should I then take over with taylor expansion for larger x? From the graph, it seems like a pretty fine lower bound.



I still feel like I should be able to make a neater argument using the convexity and decreasing properties of ln(1+1/x) and any pointers would be appreciated!



edit: Not a duplicate, the term in the denominator on the lhs has a 1/2.



Answer



Hint: Define
f1:R>0R,f1(t):=ln(1+1t)1t
and
f2:R>0R,f2(t):=ln(1+1t)1t+12
and use the intermediate value theorem twice.


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