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Can anyone tell me how to approximate the following functions?

$f_3(n) = \displaystyle\sum_{j=1}^n\sum_{k=1}^{\lfloor\frac{n}{j}\rfloor}\int_1^{\frac{n}{jk}}dx$

$f_4(n) = \displaystyle\sum_{j=1}^n\sum_{k=1}^{\lfloor\frac{n}{j}\rfloor}\sum_{l=1}^{\lfloor\frac{n}{jk}\rfloor}\int_1^{\frac{n}{jkl}}dx$

$f_5(n) = \displaystyle\sum_{j=1}^n\sum_{k=1}^{\lfloor\frac{n}{j}\rfloor}\sum_{l=1}^{\lfloor\frac{n}{jk}\rfloor}\sum_{m=1}^{\lfloor\frac{n}{jkl}\rfloor}\int_1^{\frac{n}{jklm}}dx$

and so on for $f_k(n)$ in general - and especially how to tell what the error term would be in such approximations?

I know from graphing them that they all look like quite smooth functions.

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How are the fractional summation bounds to be interpreted? – joriki Dec 5 '11 at 4:29
I'm not sure I know what you're asking, so here are some thoughts; maybe this will cover what you're asking. Sums run through the floor of their upper bounds, essentially. Integral upper limit is not floored. So, for example, in Mathematica, $f_4(n)$ would be written as F4[n_] := Sum[ Integrate[ 1, {x, 1, n/(j k l) }], {j, 1, n}, {k, 1, n/j}, {l, 1, n/(j k)}] – Nathan McKenzie Dec 5 '11 at 4:42
As far as I'm aware non-integer summation bounds are unusual; one would usually explicitly apply the floor function to get an integer bound. – joriki Dec 5 '11 at 4:57
In Mathematica, lower and upper bounds of sums are essentially truncated to integers (see the manual for details); when representing the sums as actual mathematical expressions, however, you should be adding floor/ceiling functions whenever appropriate. – J. M. Dec 5 '11 at 6:41
Is this just a matter of convention? I guess I'm not clear on what else sums with non-integer upper bounds could even mean. – Nathan McKenzie Dec 5 '11 at 6:51

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