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Math Help - Infinite series (4)

  1. #1
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    Infinite series (4)

    Prove that \sum_{n=1}^{\infty} \left(1 + \frac{1}{2} + \cdots + \frac{1}{n} \right) \frac{\sin n}{n} is convergent.
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  2. #2
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    can we use the inequality  \sum_{k=1}^{n} \frac{1}{k} < n ( the equality holds only when n = 1 ) ?

    the next step we have to go is just considering whether  \sum_{n=1}^{\infty} {\sin n} converges
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  3. #3
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    Quote Originally Posted by simplependulum View Post
    can we use the inequality  \sum_{k=1}^{n} \frac{1}{k} < n ( the equality holds only when n = 1 ) ?

    the next step we have to go is just considering whether  \sum_{n=1}^{\infty} {\sin n} converges
    the comparison test is valid for series with positive terms only. so it doesn't work here. also, even if we could use the test, it still wouldn't give us anything because \sum \sin n is divergent.
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  4. #4
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    Maybe use the integral test and we know that  \text{Si}(\infty) = \pi/2 .
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  5. #5
    MHF Contributor chisigma's Avatar
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    If we indicate with...

    H_{n} = \sum_{k=1}^{n} \frac{1}{k} (1)

    ... the 'armonic number' of index n, the 'infinite series' is...

    S= \sum_{n=1}^{\infty} H_{n} \frac{\sin n}{n} (2)

    Remembering that the generating function of armonic numbers is...

    \sum_{n=1}^{\infty} H_{n}\cdot x^{n}= - \frac {\ln (1-x)}{1-x} (3)

    ... we derive in two steps...

    \sum_{n=1}^{\infty} H_{n}\cdot x^{n-1}= - \frac{\ln (1-x)}{x\cdot (1-x)} (5)

    \sum_{n=1}^{\infty} H_{n}\cdot \frac{x^{n}}{n} = -\int_{0}^{x} \frac{\ln (1-\xi)}{\xi\cdot (1-\xi)}\cdot d\xi = \varphi(x) (6)

    Now we use the identity...

    \sin n = \frac{e^{in}-e^{-in}}{2i} (7)

    ... and (6) to arrive at the result...

    \sum_{n=1}^{\infty} H_{n}\cdot \frac{\sin n}{n} = \frac{1}{2i} \cdot \sum_{n=1}^{\infty} \frac{H_{n}}{n}\cdot (e^{in}-e^{-in}) =  \frac{1}{2i}\cdot \{\varphi(e^{i}) - \varphi(e^{-i})\} (8)

    Kind regards

    \chi \sigma
    Last edited by chisigma; May 25th 2009 at 07:27 AM. Reason: premature sending and minor details...
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  6. #6
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    Investigate the convergence of series  b_n=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n)\frac{\sin n}{n}.

    It is well known that (1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n)=\gamma+O(\frac{1}{n}), where \gamma is the Euler-Mascheroni constant.

    Obviously series {b_n} converges since b_n=\gamma\frac{\sin n}{n}+O(\frac{1}{n})\frac{\sin n}{n}, knowing \int^{\infty}_{1} \frac{\sin x}{x}dx exists.

    Now investigate the convergence of series  c_n=\ln n \frac{\sin n}{n}, leaving to check if the following integral exists
     \int^{\infty}_{1} \frac{\ln x}{x}\sin x dx. Integral by part, it is easy to know it exists.

    Therefore, a_n=(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n})\fr  ac{\sin n}{n}=b_n+c_n converges.
    Last edited by mr fantastic; September 19th 2009 at 12:47 AM. Reason: Restored original reply and fixed some latex
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  7. #7
    Moo
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    A Cute Angle Moo's Avatar
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    Luobo,

    What you did is correct.
    But it doesn't help for the present problem, since we want to prove if the series \sum a_n is convergent, not the sequence a_n.


    chisigma, how do you prove that \varphi(e^i)-\varphi(e^{-i}) converges ?
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  8. #8
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    Moo,

    I indeed is talking about series. For clarity, it is rearranged below:

    (1) Series \{b_n\} is convergent.
    \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n\right)\frac{\sin n}{n}=\gamma \sum_{n=1}^{\infty}\frac{\sin n}{n} + \sum_{n=1}^{\infty}O\left(\frac{1}{n^2}\right)\sin n { Note: 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n=\gamma+O\left(\frac{1}{n}\right) }
    The first term converges knowing \int^{\infty}_{1}\frac{\sin x}{x}\,dx is bounded, the second term converges even more obviously.

    (2) Series \{c_n\} is also convergent.
    \sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{\l n n}{n}\sin n
    It converges since \int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx is bounded.
    \left|\int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx\right|=\left|\int^{\infty}_{1}\frac{\ln x}{x}\,d\cos x\right |=\left|\int^{\infty}_{1}\frac{1-\ln x}{x^2}\cos x\,dx\right|  \leq\int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\cos x\right|\,dx\leq \int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\right|\,dx<+\infty

    (3) Series \{a_n\} is convergent
    \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\fra  c{\sin n}{n}=\sum_{n=1}^{\infty} b_n+\sum_{n=1}^{\infty} c_n
    Last edited by mr fantastic; September 19th 2009 at 12:49 AM. Reason: Restored original reply and fixed some latex
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Moo View Post
    ... chisigma, how do you prove that \varphi(e^i)-\varphi(e^{-i}) converges?...
    ... all right!... with little patience [or even using Wolfram Integrator ...] you find that...

    \varphi (x)= - \int_{0}^{x} \frac{\ln(1-\xi)}{\xi(1-\xi)} d\xi= \frac{1}{2} \ln^2 (1-x) + Li_{2} (x) (1)

    ... where Li_{2} (*) is the Polylogarithm function of order two, defined in the complex plane as...

    Li_{2} (z)= \sum_{k=1}^{\infty} \frac{z^{k}}{k^{2}} (2)

    Now the series (2) converges for |z|=1 so that \varphi (e^{i}) and \varphi (e^{-i}) both exist...

    Kind regards

    \chi \sigma
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  10. #10
    MHF Contributor chisigma's Avatar
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    Following the same procedure we arrive to demonstrate that is...

    \sum_{n=1}^{\infty} H_{n} \frac{\cos n}{n} = \frac{1}{2} \{\varphi (e^{i}) + \varphi(e^{-i})\} (1)

    These are interesting results that are not isolated, because the same procedure can be used to arrive to even better results ...

    Kind regards

    \chi \sigma
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  11. #11
    MHF Contributor chisigma's Avatar
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    An example... from the well known expression...

    \sum_{n=1}^{\infty} x^{n-1}= \frac{1}{1-x} (1)

    ... we derive...

    \sum_{n=1}^{\infty} \frac{x^{n}}{n} = \int_{0}^{x} \frac{d\xi}{1-\xi} = -\ln (1-x) (2)

    Now from (2) and the identity...

    \sin n= \frac{e^{in}-e^{-in}}{2i} (3)

    ... we derive...

    \sum_{n=1}^{\infty} \frac{\sin n}{n} = \frac{\ln (1-e^{-i}) - \ln (1-e^{i})}{2i}= \tan^{-1} \frac {\sin 1}{1-\cos 1} = 1.07079632679\dots (4)

    ... and from (2) and the identity...

     \cos n= \frac{e^{in}+e^{-in}}{2} (5)

    ... we derive...

    \sum_{n=1}^{\infty} \frac{\cos n}{n} = \frac{\ln (1-e^{-i}) + \ln (1-e^{i})}{2}= \ln \sqrt {2 (1-\cos 1)} = -.042019505825\dots (6)

    Kind regards

    \chi  \sigma
    Last edited by chisigma; September 7th 2010 at 10:51 AM. Reason: trivial error in numerical value of (4)... sorry!...
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  12. #12
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    Quote Originally Posted by luobo View Post
    Moo,

    I indeed is talking about series. For clarity, it is rearranged below:

    (1) Series \{b_n\} is convergent.
    \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\left(1+  \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n\right)\frac{\sin n}{n}=\gamma \sum_{n=1}^{\infty}\frac{\sin n}{n} + \sum_{n=1}^{\infty}O\left(\frac{1}{n^2}\right)\sin n { Note: 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n=\gamma+O\left(\frac{1}{n}\right) }
    The first term converges knowing \int^{\infty}_{1}\frac{\sin x}{x}\,dx is bounded, the second term converges even more obviously.

    (2) Series \{c_n\} is also convergent.
    \sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{\l  n n}{n}\sin n
    It converges since \int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx is bounded.
    \left|\int^{\infty}_{1} \frac{\ln x}{x}\sin x\, dx\right|=\left|\int^{\infty}_{1}\frac{\ln x}{x}\,d\cos x\right |=\left|\int^{\infty}_{1}\frac{1-\ln x}{x^2}\cos x\,dx\right|  \leq\int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\cos x\right|\,dx\leq \int^{\infty}_{1}\left |\frac{1-\ln x}{x^2}\right|\,dx<+\infty

    (3) Series \{a_n\} is convergent
    \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\left(1+  \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\fra  c{\sin n}{n}=\sum_{n=1}^{\infty} b_n+\sum_{n=1}^{\infty} c_n
    you should know that the convergence of \int^{\infty} f(x) \ dx doesn't necessarily imply that \sum^{\infty} f(n) is convergent. for example \int_0^{\infty} \sin x^2 \ dx=\sqrt{\frac{\pi}{8}} but the series \sum_{n=0}^{\infty} \sin n^2 is divergent.
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  13. #13
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    NonConAlg,

    You are right. Thanks a lot. Maybe we can switch to Abel's test. Abel's Test. Check below.

    (1) Partial sum \sum_{n=1}^{N}\sin n is bounded.
    \sum_{n=1}^{N}\sin n = Im e^i+e^{2i}+e^{3i}+...+e^{Ni} = Im \frac{e^i-e^{(N+1)i}}{1-e^i}
    \left |\sum_{n=1}^{N}\sin n\right | \leq \left|\frac{e^i-e^{(N+1)i}}{1-e^i}\right |=\frac{\left|e^i-e^{(N+1)i}\right|}{\left|1-e^i\right|} \leq \frac{\left|e^i\right|+\left|e^{(N+1)i}\right|}{\l  eft|1-e^i\right|}=\frac{2}{\left|1-e^i\right|}=\csc{\frac{1}{2}}

    (2) Series \{b_n\} is convergent.
    \sum_{n=1}^{\infty}b_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n\right)\frac{\sin n}{n}=\gamma \sum_{n=1}^{\infty}\frac{\sin n}{n} + \sum_{n=1}^{\infty}O\left(\frac{1}{n^2}\right)\sin n { Note: 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}-\ln n=\gamma+O\left(\frac{1}{n}\right) }
    The second term converges obviously. The first term also converges according to Abel's test:
    (Condition 1) Sequence \left\{\frac{1}{n}\right\} is decreasing
    (Condition 2) \lim_{n\to\infty}\frac{1}{n}=0
    (Condition 3) Partial sum \sum_{n=1}^{N}\sin n is bounded

    (3) Series \{c_n\} is also convergent.
    \sum_{n=1}^{\infty}c_n=\sum_{n=1}^{\infty}\frac{\l n n}{n}\sin n
    (Condition 1) Sequence \left\{\frac{\ln n}{n}\right\} is decreasing for n=3,4,5...
    (Condition 2) \lim_{n\to\infty}\frac{\ln n}{n}=0
    (Condition 3) Partial sum \sum_{n=1}^{N}\sin n is bounded

    (4) Series \{a_n\} is convergent
    \sum_{n=1}^{\infty}a_n=\sum_{n=1}^{\infty}\left(1+ \frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\fra  c{\sin n}{n}=\sum_{n=1}^{\infty} b_n+\sum_{n=1}^{\infty} c_n
    Last edited by mr fantastic; September 19th 2009 at 12:51 AM. Reason: Restored original reply
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  14. #14
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    The sequence a_n=\frac{H_n}n is monotonically decreasing to 0.

    Let b_k=\sum_{n=1}^k \sin n. Then b_k\sin{\textstyle\frac12}=\sum_{n=1}^k \frac{\cos(n-{\textstyle\frac12})-\cos(n+{\textstyle\frac12})}2=\frac{\cos({\textsty  le\frac12})-\cos(k+{\textstyle\frac12})}2.

    Thus |b_k|=\left|\sum_{n=1}^k \sin n\right|\leq\frac1{\sin{\textstyle\frac12}} for all k\in\mathbb N, i.e. the partial sums of \sum_{n=1}^\infty \sin n are bounded.

    By Dirichlet's test, \sum_{n=1}^\infty a_n\sin n converges.
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  15. #15
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    halbard,

    I was thinking about doing the same thing . My reference about the Abel's Test from a website appears ought to be the Dirichlet's test.

    Luobo
    Last edited by mr fantastic; September 19th 2009 at 12:52 AM. Reason: Restored original reply
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