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By looking at an integral and bounding the error?

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    What have you *tried* so far?2010-09-29
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    Yes I have tried. I got bounds on the sum by but they differ by order of \sqrt{n} which doesn't seem like a great estimate.2010-09-29
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    @blaklaybul Compare the sum with $\int_0^n\sqrt{x}dx=2n\sqrt{n}/3$.2010-09-30
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    @AD.: I suggest you check out the answers. What you said has already been said in the answers and blaklaybul's comment about sqrt(n) error is exactly about that, I believe!2010-10-01
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    @Moron: Sorry, i was a bit hasty (btw it was a nice job you did there). What I wanted to say is that it is easy to get a feel for a sum by looking at a similar integral (which are often much easier to deal with of course). Maybe too trivial comment?2010-10-01
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    @AD. Thanks! About the integral, not trivial, but OP already knew that (even before getting any answers, I think), so kind of redundant.2010-10-01
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    This sum is investigated at this [MSE link](http://math.stackexchange.com/questions/442470/).2015-01-21

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