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Math Help - Series problems

  1. #1
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    Series problems

    #1
    Find the radius of convergence and the interval of convergence of the series:
    \sum_{n=0}^{\infty}\frac{x^n}{nln(1+n)}
    I have attempted this problem as following
    By root test
    lim_{n\to\infty}\sqrt[n]{|\frac{x^n}{nln(1+n)}|}
    =lim_{n\to\infty}|x|\sqrt[n]{|\frac{1}{nln(1+n)}|}
    =lim_{n\to\infty}|x|\frac{1}{|\sqrt[n]{ln(1+n)}|}
    =\frac{|x|}{lim_{n\to\infty}\sqrt[n]{ln(1+n)}}
    =\frac{|x|}{\sqrt[n]{lim_{n\to\infty}ln(1+n)}}
    =\frac{|x|}{\sqrt[n]{lim_{n\to\infty}ln(1+n)}}
    =\frac{|x|}{\sqrt[n]{ln(lim_{n\to\infty}(1+n))}}
    =|x|

    When x=1, \sum_{n=0}^{\infty}\frac{1}{nln(1+n)} converges.
    When x=-1, \sum_{n=0}^{\infty}\frac{(-1)^n}{nln(1+n)}converges.
    Therefore, R=1 and I=[-1,1]

    Is it correct?

    #2
    Assume that the positive series \sum_{n=0}^{\infty} a_n is convergent.
    Prove that the series \sum_{n=0}^{\infty} ({a_n}^4+\frac{1}{n^2}) is also convergent.
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  2. #2
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    Re: Series problems

    For a power series, you do not have to prove convergance using root tests, because you know that the series converges within its radius of convergance, and this radius is defined whenever the limit lim a_n/a_n+1 exists. So compute the limit, then test the end points.
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  3. #3
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    Re: Series problems

    Quote Originally Posted by Chris11 View Post
    For a power series, you do not have to prove convergance using root tests, because you know that the series converges within its radius of convergance, and this radius is defined whenever the limit lim a_n/a_n+1 exists. So compute the limit, then test the end points.
    But is that true for using root test to prove it and is my computation true?
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  4. #4
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    Re: Series problems

    Quote Originally Posted by Chris11 View Post
    For a power series, you do not have to prove convergance using root tests, because you know that the series converges within its radius of convergance, and this radius is defined whenever the limit lim a_n/a_n+1 exists. So compute the limit, then test the end points.
    Surely you mean the radius of convergence is defined wherever \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1 \end{align*}...
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    Re: Series problems

    Quote Originally Posted by maoro View Post
    #2Assume that the positive series \sum_{n=0}^{\infty} a_n is convergent.
    Prove that the series \sum_{n=0}^{\infty} ({a_n}^4+\frac{1}{n^2}) is also convergent.
    As a_n\to 0 there exists n_0\in\mathbb{N} such that a_n^4\leq a_n for all n\geq n_0 . Hence, 0\leq a_n^4+\frac{1}{n^2}\leq a_n+\frac{1}{n^2} and \sum_{n=1}^{\infty}\left(a_n+\frac{1}{n^2}\right) is convergent. So, ...

    P.S. Surely there is a typo in your post: n=1 instead of n=0 .
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    Re: Series problems

    Here is another problem is need help:
    Use the ratio test to prove that the series \sum_{n=1}^{\infty} \sin{\frac{\pi}{n} diverges
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  7. #7
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    Re: Series problems

    \lim_{n\to \infty}\frac{\sin (\pi /n)}{\pi /n}=1\neq 0 , \sin (\pi/n)\geq 0 for all n , and \sum_{n=1}^{\infty}\pi/n is divergent, so the given series is divergent.
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  8. #8
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    Re: Series problems

    Quote Originally Posted by maoro View Post
    Here is another problem is need help:
    Use the ratio test to prove that the series \sum_{n=1}^{\infty} \sin{\frac{\pi}{n} diverges
    The ratio test is inconclusive because \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{\sin{ \left( \frac{\pi}{n + 1} \right) } }{ \sin{ \left( \frac{\pi}{n} \right) } } \right| = 1 \end{align*}...
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    Re: Series problems

    Quote Originally Posted by Prove It View Post
    The ratio test is inconclusive because \displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{\sin{ \left( \frac{\pi}{n + 1} \right) } }{ \sin{ \left( \frac{\pi}{n} \right) } } \right| = 1 \end{align*}...
    ya, i also got this result and can't prove the desired divergence.
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  10. #10
    MHF Contributor FernandoRevilla's Avatar
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    Re: Series problems

    Quote Originally Posted by maoro View Post
    ya, i also got this result and can't prove the desired divergence.
    For that reason I applied the Limit Comparison Test: if \sum a_n and \sum b_n are series with positive terms and the limit \lim_{n\to \infty}\frac{a_n}{b_n} exists and is nonzero, then either both \sum a_n and \sum b_n converge, or else both diverge.
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  11. #11
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    Re: Series problems

    @proveit I was just telling him what the radius was.
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