# Tag Info

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### asymptotic of a determinant

The matrix $M_n$ is the Hilbert matrix $H_n = \left(\dfrac{1}{i+j+1}\right)_{0 \le i,j \le n-1}$ but flipped horizontally. Hence, $|\det(M_n)| = |\det(H_n)|$ for all $n$. From the Wikipedia article on ...
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### Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?

We have \begin{align*} \prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ &...
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### Asymptotic behavior of integral with Laplace's method

Elaborating on Maxim's comment. If $v=xT$, then \begin{align*} I(n)&=\int_0^1\!\! {\int_0^1 {e^{ - nxT} \sqrt {1 - (1 - \sqrt x + \sqrt {xT} )^2 } dT} dx} \\& = \int_0^1\!\! {\int_0^x {e^{ - ...
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### sequence such that $x_{n+1}=n(x_n-n)$

By some experimentation from the starting point $x_n=an+b+y_n$ and reporting in the equation I quickly found that $x_n=n+1+2y_n$ lead to the following simplification $$y_{n+1}=ny_n-1$$ So we have ...
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### Asymptotics of a two dimensional integral

It turns out that @Gary 's suggestion to reverse the order of integration is a fruitful first step because the integral over $\epsilon$ is analytically tractable. Exchanging the order results in the ...
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### Bounding a convergent sum

For each $n$, let $f_n(x) = \frac{\exp(-n/2^x)}{2^x x}$. Then we can show that(1),(2) $$\left| \sum_{i=1}^{n} f_n(i) - \int_{1}^{n} f_n(x) \, \mathrm{d}x \right| \leq \max_{1\leq x \leq n} f_n(x).$$ ...

### Is this a sound line of reasoning to conclude that $\sqrt[n]{n!} \sim \frac{n}{e}$?

You may just use the following Lemma if $\{a_n\}_{n\geq 1}$ is a sequence of positive numbers such that $\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=L$, then $\lim_{n\to +\infty}\sqrt[n]{a_n}=L$. Apply ...

### Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$?

Asymptotically, $$g(n) = \prod_{2 < p < n}^{p \text{ prime}} 1-\frac2p \sim \frac{4C_2e^{-2\gamma}}{(\log n)^2} \approx \frac{0.83244}{(\log n)^2}$$ And presumably your expression is simply the ...
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1 vote

### Expansion of logarithm with little o

When you have on $o(z)$ in a development it means that anything smaller than order $z$ is simply ignored. When $z\to 0$ we have to ignore any terms in $z^a$ where $a>1$, in particular $z^2$ and of ...
1 vote

### Expansion of logarithm with little o

You were too aggressive with your $o$s, and confused an $O(|z+o(z)|^2)$ term with $o(|z+o(z)|^2)$. You could have instead expanded as: $$z+o(z)-\frac{1}{2}(z+o(z))^2+o(|z+o(z)|^2)$$ We can clean this ...

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