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Limits of sequences
05-22-2010, 10:24 AM
Post: #1
Limits of sequences
(1) \(\displaystyle{\lim_{n \to \infty}} \sin(2\pi e \cdot n!)\)
(2) \(\displaystyle{\lim_{n \to \infty}} \sum_{k=1}^n \frac{(-1)^{k-1}}{(k+(-1)^{k-1})^p}\) where \(p \in (0,1)\)
(3) Does \(\{|\sin n|^{\frac{1}{n}}\}\) converge?
(4) \(\displaystyle{\lim_{n \to \infty} (\sqrt{n^2+1}+\sqrt{n^2+2}+\cdots+\sqrt{n^2+n} -n^2 - \frac{n}{4})}\)
(5) \(\displaystyle{(a_n > 0 \quad (\forall n), \quad \lim_{n \to \infty} \frac{a_n a_{n+2}}{a_{n+1}^2} = \lambda )\Rightarrow (\lim_{n \to \infty} a_n^{\frac{1}{n(n+1)}} = \sqrt{\lambda})}\)
(6) $\displaystyle{ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n+k}}$
(7) $\displaystyle{\lim_{n \to \infty} \frac{\sum_{k=1}^n \frac{1}{k}}{\ln n}}$
(8) $\displaystyle{\lim_{n\to \infty} \frac{\frac{1}{n+1}\sum_{k=0}^n (a+bk)}{\prod_{k=0}^n (a+bk)^{\frac{1}{n+1}}}}$
(9) $\displaystyle{\lim_{n \to \infty} \frac{\sum_{k=1}^n x_k}{\sum_{k=1}^n \frac{1}{k}}}$, where $\displaystyle{x_1 = 1,\, x_{n+1} = \frac{x_n^2 + 2}{n+1}}$
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05-26-2010, 07:46 AM
Post: #2
RE: Limits of sequences: (1)
(1) $ \displaystyle{n! e= n! \sum_{k=0}^{\infty} \frac{1}{k!} = n! \sum_{k=0}^n \frac{1}{k!} + R_n}$
\(\quad\quad\)where $\displaystyle{R_n = \frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\cdots < \sum_{i=1}^{\infty} \frac{1}{(n+1)^i} = \frac{1}{n} \to 0 \; (n \to \infty)}$
\(\quad\quad\)So $\displaystyle{\lim_{n\to \infty}} \sin(2\pi n! e) = \displaystyle{\lim_{n\to \infty}} \sin (2\pi R_n) = 0$
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08-26-2010, 05:25 PM
Post: #3
RE: Limits of sequences: (2)
(2) \(\displaystyle{\sum_{k=1}^n \frac{(-1)^{k-1}}{(k+(-1)^{k-1})^p} = - \sum_{k=1}^n \frac{(-1)^{k-1}}{k^p} + \frac{(1-(-1)^n)(-1)^{n-1}}{2(n+(-1)^{n-1})^p} \to -\eta(p) = (2^{1-p}-1)\zeta(p)}\)
$\displaystyle{\eta(p) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^p}=\frac{1}{1^p}-\frac{1}{2^p}+\frac{1}{3^p}-\cdots} $
\(\quad \quad\)http://en.wikipedia.org/wiki/Dirichlet_eta_function
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08-28-2010, 10:30 AM
Post: #4
RE: Limits of sequences: (3)
(3) From "Pi, A Source Book" (3rd Ed. page 309) we have
\(\quad \quad\)$|\pi - p/q| > q^{-42} \; (\forall p,q \in \mathbb{N}, \; p,q > 2)$
\(\quad \quad\)For $n > 10$, let $q_n$ be the nearest integer to $n/\pi$, then $\pi/2 >|n -\pi q_n| > q_n^{-41} > n^{-41}$
\(\quad \quad\)and so $|\sin n| = |\sin (n -\pi q_n)| = \sin |n -\pi q_n| > |n -\pi q_n|/n > n^{-42}$

Therefore $1 > |\sin n|^{1/n} > n^{-42/n} \to 1 \; (n \to \infty) $
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09-10-2010, 07:28 AM
Post: #5
RE: Limits of sequences: (4)
(4) From \(\sqrt{1+x} = 1+x/2 - x^2 /8 + O(x^3), \quad (|x| < 1)\), $\sqrt{n^2 + k} = n\displaystyle{(1+\frac{k}{2n^2}-\frac{k^2}{8n^4} + O(n^{-3})) }$
\(\quad\quad\)thus $\sum_{k=1}^n \sqrt{n^2+k} - n^2- n/4 = \displaystyle{n^2+\frac{n(n+1)}{4n}-(n^2+n/4)-\frac{1}{8}\sum_{k=1}^n \left(\frac{k}{n} \right)^2 \frac{1}{n} + O(1/n)}$
\(\quad\quad\) and \(\displaystyle{\lim_{n \to \infty} \left(\sum_{k=1}^n \sqrt{n^2+k} -n^2 -n/4\right)}= 1/4 -\frac{1}{8}\int_0^1 x^2 dx = \frac{5}{24}\)
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11-30-2010, 06:07 PM
Post: #6
RE: Limits of sequences: (5)
(5) \(\displaystyle{(a_n > 0 \quad (\forall n), \quad \lim_{n \to \infty} \frac{a_n a_{n+2}}{a_{n+1}^2} = \lambda )\Rightarrow (\lim_{n \to \infty} a_n^{\frac{1}{n(n+1)}} = \sqrt{\lambda})}\)

Prove: By Stolz, $\displaystyle{\lim_{n \to \infty} \ln \sqrt[n(n+1)]{a_n} = \lim_{n \to \infty}\frac{\ln a_n}{n(n+1)}=\lim_{n \to \infty} \frac{\ln \frac{a_{n+1}}{a_n}}{2(n+1)}}$
$\quad\quad \displaystyle{=\lim_{n \to \infty}\frac{\ln\frac{a_{n+1}}{a_{n+1}}-\ln \frac{a_{n+1}}{a_n}}{2(n+2)-2(n+1)} =\lim_{n \to \infty}\frac{\ln \frac{a_n a_{n+2}}{a_{n+1}^2}}{2} =\ln \sqrt{\lambda}}$
$\quad\quad$ Therefore $\displaystyle{\lim_{n \to \infty} \sqrt[n(n+1)]{a_n} = \sqrt{\lambda}} \quad \square$
   
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12-18-2010, 01:25 PM
Post: #7
RE: Limits of sequences: (6)
(6) $\displaystyle{\lim_{n\to\infty} \sum_{k=1}^n \frac{1}{n+k}=\lim_{n\to\infty}\sum_{k=1}^n \frac{1}{1+\frac{k}{n}}\frac{1}{n} = \int_0^1 \frac{1}{1+x}dx = \ln 2}$
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12-18-2010, 08:01 PM
Post: #8
RE: Limits of sequences: (7)
(7) $\displaystyle{\lim_{n\to\infty}\frac{\sum_{k=1}^n \frac{1}{k}}{\ln n}=\overset{stolz}{=}\lim_{n\to\infty}\frac{1/(n+1)}{\ln (1+\frac{1}{n})} =\lim_{n\to\infty}\frac{n/(n+1)}{\ln (1+\frac{1}{n})^n}= 1}$
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12-25-2010, 12:29 AM
Post: #9
RE: Limits of sequences: (8)
(8) We have the following calculation
$\quad\quad\displaystyle{\ln \left(\prod_{k=1}^n (a+bk)^{\frac{1}{n+1}} \right )=\frac{1}{n+1}\sum_{k=0}^n \ln(a+bk)}$
$\quad\quad \displaystyle{\sum_{k=0}^n \ln(a+bk) < \int_0^{n+1} \ln(a+bx) dx <\sum_{k=1}^{n+1}\ln(a+bk)}$
$\quad\quad \displaystyle{0<\int_0^{n+1}\ln(a+bx)dx-\sum_{k=0}^n \ln(a+bk)<\ln\left(1+\frac{b}{a}(n+1)\right)}$
$\quad\quad \displaystyle{\int \ln(a+bx)dx = x\ln(a+bx)+\frac{a}{b}\ln\left(\frac{a}{b}+x\right)-x},\quad$ thus
$\quad\quad \displaystyle{\ln \left(\prod_{k=0}^n (a+bk)^{\frac{1}{n+1}} \right )=\frac{1}{n+1}\sum_{k=0}^n \ln(a+bk) =}$
$\quad\quad\quad =\displaystyle{\frac{(n+1)\left(\ln(a+b(n+1))-1 \right )+\frac{a}{b}\ln\left(1+\frac{b}{a}(n+1) \right )+O(\ln n)}{n+1}}$

$\quad\quad\quad = \ln\left(a+b(n+1)\right)-1+o(n)$

$\quad\quad$Also $ \displaystyle{\sum_{k=0}^n (a+bk) = \frac{1}{2} (2a +bn)(n+1)}$, therefore
$\quad\quad \displaystyle{\lim_{n\to\infty} \frac{\frac{1}{n} \sum_{k=0}^n (a+bk)}{\prod_{k=0}^n (a+bk)^{\frac{1}{n+1}}}=\lim_{n\to\infty} \frac{\frac{n+1}{2n}(2a+bn)}{(a+b(n+1))e^{-1} e^{o(n)}}=\frac{e}{2}}$
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09-07-2012, 05:51 PM
Post: #10
RE: Limits of sequences (9)
Since $\displaystyle{x_1 < 2,\; x_n <2 \implies x_{n+1} < \frac{2^2 + 2}{2+1} = 2\,(n\ge 2)}$ thus $x_n < 2\,$ and $0 < x_n < \frac{6}{n+1} \to 0$

Now $\displaystyle{\frac{x_n}{\frac{1}{n}} = n x_n= x_{n-1}^2 + 2 \to 0}$ so Cesaro-Stolz theorem gives $\displaystyle{\lim_{n\to\infty}\frac{\sum_{k=1}^n x_k}{\sum_{k=1}^n \frac{1}{k}} = \lim_{n\to\infty}\frac{x_n}{\frac{1}{n}}= 2}$
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