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steven_zhang123

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steven_zhang123

#1 h Oct 31, 2024, 1:47 PM

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Show that $\lim_{n \to \infty} \sum_{k=1}^{n} (\sqrt{1+\frac{k}{n^{2} } } -1)=\frac{1}{4}$ .

This post has been edited 2 times. Last edited by steven_zhang123, Oct 31, 2024, 2:31 PM

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wolfqz

#2 h Oct 31, 2024, 2:22 PM

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steven_zhang123 wrote:

Show that $\lim_{x \to +\infty} \sum_{k=1}^{n} (\sqrt{1+\frac{k}{n^{2} } } -1)=\frac{1}{4}$ .

Did you mean $\lim_{n \to \infty} \sum_{k=1}^{n} \left(\sqrt{1+\frac{k}{n^{2} } } -1\right)=\frac{1}{4}$ ?

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steven_zhang123

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steven_zhang123

#3 h Oct 31, 2024, 2:31 PM

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Sure. I think maybe I should change it.

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#4 h Oct 31, 2024, 3:25 PM • 1 Y

Y by Svyatoslav

First notice
$\sum_{k=0}^n\frac{k}{n^2}=\frac{n+1}{2n}\overset{n\to\infty}{\to}\frac{1}{2}$ So we subtract our given sum from $\sum_{k=0}^n\frac{k}{2n^2}$ and evaluate:
$0<\sum_{k=0}^n\left(1+\frac{k}{2n^2}-\sqrt{1+\frac{k}{n^2}}\right)=\sum_{k=0}^n\frac{\frac{k^2}{4n^4}}{\sqrt{1+\frac{k}{n^2}}+1+\frac{k}{2n^2}}<\sum_{k=0}^n\frac{k^2}{n^4}=O\left(\frac{n^3}{n^4}\right)=O\left(\frac{1}{n}\right)\overset{n\to\infty}{\to}0$ $\blacksquare$

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#5 h Oct 31, 2024, 3:44 PM • 3 Y

Y by teomihai, programjames1, Svyatoslav

By multiplying by the conjugate on numerator and denominator, the sum is equal to
$\sum_{k=1}^n \frac{k}{n^2\left(\sqrt{1+\frac{k}{n^2}}+1\right)}$ Note that for $k\in [0,n]$ , $\frac{1}{1+\sqrt{1+\frac{1}{n}}}\leq\frac{1}{\sqrt{1+\frac{k}{n^2}}+1}\leq\frac{1}{2}$ and both sides approach $\frac{1}{2}$ as $n\to\infty$ .

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#6 h Oct 31, 2024, 4:50 PM • 1 Y

Y by Svyatoslav

natmath wrote:

By multiplying by the conjugate on numerator and denominator, the sum is equal to
$\sum_{k=1}^n \frac{k}{n^2\left(\sqrt{1+\frac{k}{n^2}}+1\right)}$ Note that for $k\in [0,n]$ , $\frac{1}{1+\sqrt{1+\frac{1}{n}}}\leq\frac{1}{\sqrt{1+\frac{k}{n^2}}+1}\leq\frac{1}{2}$ and both sides approach $\frac{1}{2}$ as $n\to\infty$ .

Yes,nice! or shortly :

$\sum_{k=1}^n \frac{k}{n^2\left(\sqrt{1+\frac{n}{n^2}}+1\right)}\leq{\sum_{k=1}^n \frac{k}{n^2\left(\sqrt{1+\frac{k}{n^2}}+1\right)}}\leq{\sum_{k=1}^n \frac{k}{n^2\left(\sqrt{1+\frac{1}{n^2}}+1\right)}}$ result with sandwich criteria , limit is $\frac{1}{4}$ .
Analogously it is demonstrated that :
$\lim_{n \to \infty} \sum_{k=1}^{n} (\sqrt [p]{1+\frac{k^p}{n^{p+1} } } -1)=\frac{1}{p^2}$ for any positiv integer $p\geq{2}$ .
Finally applying STOLZ -CESARO criteria.

This post has been edited 2 times. Last edited by teomihai, Oct 31, 2024, 5:03 PM

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#7 h Nov 26, 2024, 2:46 PM • 1 Y

Y by teomihai

We can do this easily by Teplitz's theorem.

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Kath_revenge_krr

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Kath_revenge_krr

#8 h Mar 13, 2025, 5:49 PM

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Tepltiz theroem and expand

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#10 h Yesterday at 8:19 AM • 1 Y

Y by teomihai

Let's consider
$S(\alpha, n)=\sum_{k=1}^n\left(1+\frac k{n^2}\right)^{-\alpha}, \alpha\in R$ We note that $S(\alpha, n)$ is the analytical function of $\alpha$ - does not have branch points at any real $\alpha$ and fixed integer $n$ . Therefore, we can use the analytical continuation to find $S\Big(\alpha =-\frac12, n\Big)$
Choosing $\alpha >1$ , we can use a simple representation of gamma-function:
$S(\alpha, n)=\frac1{\Gamma(\alpha)}\sum_{k=0}^n\int_0^\infty t^{\alpha-1}e^{-t(1+\frac k{n^2})}dt, \,\alpha>1$ Changing the order of summation and integration,
$=\frac1{\Gamma(\alpha)}\int_0^\infty t^{\alpha-1}e^{-t}\frac{1-e^{-\frac t n}}{e^\frac t{n^2}-1}dt=\frac1{\Gamma(\alpha)}\left(I_1+I_2\right)\qquad\qquad(0)$ where
$I_1=n^2\int_0^\infty t^{\alpha-2}\frac{\frac t{n^2}}{e^\frac t{n^2}-1}e^{-t}dt$ Using $\frac x{e^x-1}=\sum_{k=0}^\infty B_k\frac{x^k}{k!}$ , where $B_k$ denotes Bernoulli's numbers, we find the asymptotics of $I_1$ at $n\to\infty$
$I_1\sim n^2\int_0^\infty t^{\alpha-1}e^{-t}\left(\sum_{k=0}^\infty\frac{B_k}{k!}\left(\frac t{n^2}\right)^k\right)dt\sim n^2\sum_{k=0}^\infty\frac{\Gamma(\alpha-1+k)B_k}{k!}\,\frac1{n^{2k}}\qquad(1)$ In the same way,
$I_2\sim n^2\sum_{k=0}^\infty\frac{\Gamma(\alpha-1+k)B_k}{k!}\,\left(1+\frac 1n\right)^{1-\alpha-k}\frac1{n^{2k}}\qquad\qquad\qquad\qquad(2)$ Putting (1) and (2) into (0), we find the most general asymptotics of $S(\alpha, n)$ at $\alpha\in R$
$\boxed{\,\,S(\alpha, n)\sim\frac{n^2}{\Gamma(\alpha)}\sum_{k=0}^\infty\frac{B_k\,\Gamma(\alpha-1-k)}{k!\,n^{2k}}\left(1-\left(1+\frac1n\right)^{1-\alpha-k}\right)\,\,}$ Now, choosing $\alpha=-\frac12$
$S\Big(-\frac12, n \Big)\sim-\,\frac{2n^2\,B_0}3\left(1-\Big(1+\frac1n\Big)^{\frac32}\right)+B_1\left(1-\sqrt{1+\frac1n}\right)-\frac{n^2}{2\sqrt\pi}\sum_{k=0}^\infty\frac{B_{k+2}}{(k+2)!}\,\frac{\Gamma\big(k+\frac12\big)}{n^{4+2k}}\left(1-\frac1{\big(1+\frac1n\big)^{k+\frac12}}\right)$ $S\Big(-\frac12, n \Big)-n\sim-\,\frac{2n^2}3\left(1-\Big(1+\frac1n\Big)^{\frac32}\right)-\,n-\,\frac12\left(1-\sqrt{1+\frac1n}\right)-\,\frac1{2\sqrt\pi n^2}\sum_{k=0}^\infty\frac{B_{k+2}}{(k+2)!}\,\frac{\Gamma\big(k+\frac12\big)}{n^{2k}}\left(1-\frac1{\big(1+\frac1n\big)^{k+\frac12}}\right)$ $S\Big(-\frac12, n \Big)-n=\frac14+\frac 5{24 n}-\frac 3{64 n^2}+O\Big(\frac 1{n^3}\Big)$ Numeric check (WA): here and here

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