∞∑n=11n2≤1+∫∞11x2dx
I'm having trouble figuring out why the inequality above is true. I understand the following inequality:
∫∞11x2dx≤∞∑n=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:
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
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