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Determine the convergence of series
07-29-2010, 06:10 PM
Post: #1
Determine the convergence of series
(1) $\displaystyle{\sum_{n=1}^{\infty}\frac{1}{1+\sqrt{2}+\sqrt[3]{3}+\cdots +\sqrt[n]{n}}}$
(2) $\displaystyle{\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sin(\log n)}{n^{\alpha}}\;(\alpha\in\mathbb{R})}$
(3) \(\displaystyle{\sum_{n=1}^{\infty} \ln\left(n\sin \frac{1}{n}\right)}\)
(4) \(\displaystyle{\sum_{n=1}^{\infty} \displaystyle{\frac{1}{n^{2-\sin n}}}}\)
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09-03-2012, 07:15 PM
Post: #2
RE: Decide convergence of series
Sol. (1) Since $\sqrt[n]{n} \to 1$, We have $\displaystyle{\frac{\sum_{k=1}^n\sqrt[k]{k}}{n} \to 1}$ therefore $\displaystyle{\frac{1}{1+\sqrt{2}+\cdots +\sqrt[n]{n}} \sim O\left(\frac{1}{n}\right)}$
$\quad\quad\quad$ as $n \to \infty$. So the given series diverges as $\sum \frac{1}{n}$ diverges.
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09-05-2012, 09:39 PM
Post: #3
RE: Determine the convergence of series
Sol. (2) Assume that $\alpha>0.$

By the Mean Value Theorem, $|\sin(\log(n+1))-\sin(\log(n))|\le \frac1n.$

Thus, $\left|\sum_{k=1}^n(-1)^k\sin(\log(k))\right| \le\sum_{j=1}^{\lfloor n/2\rfloor} |\sin(\log(2j))-\sin(\log(2j-1))|+1$

$\le\sum_{j=1}^{\lfloor n/2\rfloor}\frac1{2j-1}+1\le C(1+\log n)$ for some $C>0.$

We could work out $C$ more explicitly but we don't have to.

Now we employ Abel's summation by parts identity.

Let $A_n=\sum_{k=1}^n(-1)^k\sin(\log(k)).$

$\sum_{k=1}^n\frac{(-1)^k\sin(\log(k))}{k^{\alpha}}=\frac{A_n}{n^{\alpha}} +\sum_{k=1}^{n-1}A_k\left(\frac1{k^{\alpha}}-\frac1{(k-1)^{\alpha}}\right)$

Now take the limit as $n\to\infty.$ Since $A_n$ only grows at a logarithmic rate, we have that $\frac{A_n}{n^{\alpha}}\to0.$ What remains is the sum, and that sum converges absolutely:

$\sum_{k=1}^{\infty}\left|A_k\left(\frac1{k^{\alpha}}-\frac1{(k-1)^{\alpha}}\right) \right|\le C\sum_{k=1}^{\infty}\frac{1+\log k}{k^{\alpha+1}}\le C'\sum_{k=1}^ {\infty}\frac1{k^{\beta}}<\infty$

for some $1<\beta<1+\alpha$.

If $\alpha\le0,$ the original series diverges because its terms fail to go to zero.

src
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09-05-2012, 10:21 PM
Post: #4
Sol. (3)
$\quad\quad\quad$ Since $\;\displaystyle{\lim_{x\to 0} \frac{\log\left(\frac{\sin x}{x} \right )}{x^2} = \lim_{x\to 0} \frac{x}{\sin x} \frac{x\cos x-\sin x}{2x^3}=\lim_{x\to 0}\frac{-x\sin x}{6x^2}= -\frac{1}{6}}$,
$\quad\quad\quad$ We have $\;\displaystyle{\lim_{n\to\infty} n^2 \log\left( n\sin \frac{1}{n} \right ) = -\frac{1}{6}}.\quad$ So $\displaystyle{\sum_{n=1}^{\infty} \log\left(n\sin \frac{1}{n} \right )}$ converges.
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