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I have some trouble understanding the following number-theoretical estimation:

$$\sum_{k\le \sqrt{n}} (1-k^2/n)^{1+o_n(1)}=n^{1/2+o(1)} \ (n\to\infty),$$ where $o_n(1)$ denotes a $o(1)$ function depending only on $n$.

The side $\ge$ is obvious, but the other one gives me trouble.

Indeed, without the $f(k)=o_n(1)$ term, I'd try to apply integral comparison or such, but having this unknown function I can't do that.

My idea was to make it disappear in an inequality, but since I don't known anything about the sign of $f$ I'm blocked. For example, if $f\ge 0$, I could argue that $$(1-k^2/n)^{f(k)}\le 1,$$ but it is not necessarily true.

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up vote 2 down vote accepted

Divide both sides by $n^{1/2}$: $$ \lim_{n\to\infty}\sum_{k\le\sqrt{n}}(1-k^2/n)^{1+o(1)}\frac1{\sqrt{n}}=\int_0^1(1-x^2)\,\mathrm{d}x=\frac23 $$ since the left hand side is a Riemann sum with partition size $\frac1{\sqrt{n}}$. The exponent tends to $1$ as $n\to\infty$.

Therefore, $$ \sum_{k\le\sqrt{n}}(1-k^2/n)^{1+o(1)}\frac1{\sqrt{n}}\sim\frac23n^{1/2}=n^{1/2+\frac{\log(2/3)}{\log(n)}}=n^{1/2+o(1)} $$

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Thank you. Hum, ok with the fact that $\sum_{k\le\sqrt{n}} (1-k^2/n)^{1+o(1)}\sum n^{1/2+o(1)}$, but how do you see that $$\left|\sum_{k\le\sqrt{n}} (1-k^2/n)-\sum_{k\le\sqrt{n}} (1-k^2/n)^{1+o(1)}\right|$$ is arbitrarily small, without knowing anything about the $o(1)$ ? – Klaus Jan 26 '13 at 19:04
We do know that $\lim\limits_{n\to\infty}o(1)=0$. Suppose that we know that for sufficiently large $n$, $|o(1)|<\epsilon$. Then $$ \int_0^1(1-x^2)^{1+\epsilon}\mathrm{d}x\le\lim_{n\to\infty}\sum_{k\le\sqrt{n}}(1‌​-k^2/n)^{1+o(1)}\frac1{\sqrt{n}}\le\int_0^1(1-x^2)^{1-\epsilon}\mathrm{d}x $$ Since $\epsilon$ was arbitrary, and each bounding integral tends to $$\int_0^1(1-x^2)\,\mathrm{d}x$$ we are okay. – robjohn Jan 26 '13 at 19:29

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