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I don't know how to find tight (aka asymptotic) bounds for a function. Consider the function $$f(n)=\sum_{k=1}^nk^r$$

How would I find tight upper/lower bounds for this. Please help :-(

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Do you mean $f(n)$, or perhaps $f(r)$? There is no $x$ on the right-hand-side, so $f(x)$ would be constant. – robjohn May 14 '12 at 11:05
Sorry, I meant f(n) – canton May 14 '12 at 11:07
up vote 1 down vote accepted

Some trivial estimates: first of all, let's use $f(n)$ instead of $f(x)$ since there are no $x$'s so far. Let us Denote $a(x) = x^r$ for $x\geq 1$. Then your formula is $$ f(n):=\sum\limits_{k=1}^n a(k). $$ In the case when $r\geq 0$ we obtain that $a(x)$ is a non-decreasing function and hence $$ \int\limits_{n-1}^na(x)\mathrm dx\leq a(n)\leq\int\limits_n^{n+1}a(x)\mathrm dx\leq a(n+1) $$ for all $n\geq 2$. As a result, $$ a(1)+\int\limits_1^n a(x)\mathrm dx\leq\sum\limits_{k=1}^n a(k)\leq a(1)+\int\limits_1^{n+1} a(x)\mathrm dx $$ where $a(1) = 1$. Clearly, for integrals you have $$ \int\limits_1^n x^r\mathrm dx = \frac{n^{r+1}-1}{r+1} $$ so $$ 1+\frac{n^{r+1}-1}{r+1}\leq f(n)\leq \frac{(n+1)^{r+1}-1}{r+1} $$ for all $n\geq 1$.

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Okay, I'm following your answer but it seems like a puzzle I wouldn't put together on my own... But I have a question. So for the upper limit I can say that sum(k^r)<=sum(n^r) and theres an easy upper limit. Is there also a way to do this for the lower limit? – canton May 14 '12 at 10:51

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