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Math Help - Infinite Series

  1. #1
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    Infinite Series

    Need help with the following:

    Evaluate \sum_{n=0}^{\infty }\sin \left( {\frac {1}{2^n}} \right) \cos \left( {\frac {3}{2^n}} \right)

    I recognise that the terms inside the sin and cos functions are the terms of a geometric progression with first term 1 and 3 respectively and common ratio of 0.5. But when it comes to evaluating the series with the trigonometric functions applied i'm quite lost. Is there some kind of general principle to adhere to to tackle problems of this nature?

    Determine if  \sum _{n=0}^{\infty } \left( {\frac {3n}{4\,n+1}} \right) ^{2\,n} is convergent or divergent.

    I'm abit confused as to which test to use for convergence/divergence for this particular problem. I'm thinking either limit or limit comparison, but I still can't seem to work it out. Would appreciate any help you guys might be able to offer.

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Hweengee View Post
    Need help with the following:

    Evaluate \sum_{n=0}^{\infty }\sin \left( {\frac {1}{2^n}} \right) \cos \left( {\frac {3}{2^n}} \right)
    Remember that 2\sin A\cos B = \sin (A+B) + \sin (A-B)

    Therefore,
    2\sum_{n=0}^{\infty} \sin \left( {\frac {1}{2^n}} \right) \cos \left( {\frac {3}{2^n}} \right) = \sum_{n=0}^{\infty} \sin \left( \frac{1}{2^{n-2}} \right) - \sin \left( \frac{1}{2^{n-1}}\right) = \sin 4
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  3. #3
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    Thanks! And would it be correct to say that for the 2nd series,
     <br />
\sum _{n=0}^{\infty } \left( {\frac {3n}{4\,n+1}} \right) ^{2\,n}<br />
    is convergent

    since
     <br />
\lim _{n\rightarrow \infty}\left({\frac {3n}{4\,n+1}}\right)^{2\,n}=0<br />
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  4. #4
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    Quote Originally Posted by Hweengee View Post
     <br />
\lim _{n\rightarrow \infty}\left({\frac {3n}{4\,n+1}}\right)^{2\,n}=0<br />
    Use the root test. Notice we get  \left( \frac{3n}{4n+1} \right)^2 \to \left( \frac{3}{4} \right)^2 < 1.
    Thus, it is convergent.
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Hweengee View Post
    Thanks! And would it be correct to say that for the 2nd series,
     <br />
\sum _{n=0}^{\infty } \left( {\frac {3n}{4\,n+1}} \right) ^{2\,n}<br />
    is convergent

    since
     <br />
\lim _{n\rightarrow \infty}\left({\frac {3n}{4\,n+1}}\right)^{2\,n}=0<br />
    Or you could alternatively say that since

    \lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{3n}{4n+1}\right)^{2n}\right|}=\  left|\frac{9}{16}\right|<1

    Therefore it is convergent by the root test

    EDIT: sorry...I noticed that hacker already did it this way...you could also see that

    0\leq\left(\frac{3n}{4n+1}\right)^{2n}\leq\left(\f  rac{3n}{4n}\right)^{2n}=\left(\frac{9}{16}\right)^  n

    A convergent geometric series
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  6. #6
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    Quote Originally Posted by Mathstud28 View Post
    0\leq\left(\frac{3n}{4n+1}\right)^{2n}\leq\left(\f  rac{3n}{4n}\right)^{2n}=\left(\frac{9}{16}\right)^  n

    A convergent geometric series
    Yes this way is nicer.
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  7. #7
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    very nice, thanks guys.
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