Friday, 18 September 2015

integration - Why is the inequality suminftyn=1frac1n2leq1+intinfty1frac1x2 true?

n=11n21+11x2dx



I'm having trouble figuring out why the inequality above is true. I understand the following inequality:



11x2dxn=11n2



It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:




enter image description here



So let's say I rewrite n=11n2 as 1+n=21n2 since they are equivalent.



Why is that less than 1+11x2dx?



If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.



My claims come specifically from page 60 of this webpage from Dartmouth

No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...