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Math Help - seriously hitting my boiling point on these problems please help

  1. #1
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    seriously hitting my boiling point on these problems please help

    The two problems require to approximate the series to four decimal points. both are assumed to be convergent
    1) sin2n/n5 from 1 to inifinity
    I have no idea what to do because Its not an alternating series and therefore I cant use the alternating series est thm and I dont have advanced enough learning to know how to take the integral for the intergral approx test.

    2)(-1)n-1(n)/(n4+1) from 1 to infinity

    please help
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  2. #2
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    Re: seriously hitting my boiling point on these problems please help

    "Both are assumed to be convergent" - then why do you think you would need to use the Alternating Series test or Integral test to check its convergence?

    If you are wanting to find a series to represent $\displaystyle \begin{align*} \frac{\sin^2{(n)}}{n^5} \end{align*}$, I'd write $\displaystyle \begin{align*} \sin^2{(n)} = \frac{1}{2} - \frac{1}{2}\cos{(2n)} \end{align*}$ and use the Maclaurin series for $\displaystyle \begin{align*} \cos{(2n)} \end{align*}$.
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    Re: seriously hitting my boiling point on these problems please help

    sorry i didnt specify not the alternating series test for convergence, i'm talking about the alternating series estimation therom (ie. |S-Sn| < bn+1)
    and the intergral estimation sum:
    integral(n+1 to infinity) f(x) < Rn < integral (n to infinity) f(x)

    where S= Value the series converges to
    Sn= nth Partial Sum
    Rn= Remainder or 'error'
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    Re: seriously hitting my boiling point on these problems please help

    Hi,
    In your statement of an integral test, who is function f? Answer: f is a decreasing non-negative function on say $[1,\infty)$ with $f(n)=a_n$ and $\int_1^\infty f(x)\,dx$ is finite. Here also $R_n=\sum_{k=n+1}^\infty a_n$.

    Your problem:
    $$b_n={sin^2(n)\over n^5}\text{. Then }R_n=|\sum_{k=n+1}^\infty b_n|\leq\sum_{k=n+1}^\infty {1\over k^5}$$

    So let $f(x)={1\over x^5}$. Use this to find an error bound for $\sum_{k=n+1}^\infty {1\over k^5}$ and hence for $R_n$

    Your second series is an alternating series with the terms decreasing.
    Last edited by johng; March 14th 2014 at 09:48 PM.
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    Re: seriously hitting my boiling point on these problems please help

    Quote Originally Posted by johng View Post
    Hi,
    In your statement of an integral test, who is function f? Answer: f is a decreasing non-negative function on say $[1,\infty)$ with $f(n)=a_n$ and $\int_1^\infty f(x)\,dx$ is finite. Here also $R_n=\sum_{k=n+1}^\infty a_n$.

    Your problem:
    $$b_n={sin^2(n)\over n^5}\text{. Then }R_n=|\sum_{k=n+1}^\infty b_n|\leq\sum_{k=n+1}^\infty {1\over k^5}$$

    So let $f(x)={1\over x^5}$. Use this to find an error bound for $\sum_{k=n+1}^\infty {1\over k^5}$ and hence for $R_n$

    Your second series is an alternating series with the terms decreasing.
    I thought you were unable to apply the alternating series estimation thm to a none alternating series though??
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    Re: seriously hitting my boiling point on these problems please help

    Quote Originally Posted by smoez View Post
    I thought you were unable to apply the alternating series estimation thm to a none alternating series though??
    \sum_{n=1}^\infty \frac{(-1)^{n-1}n}{n^4+1} is an alternating series - whose terms decrease in absolute value. But you are correct - you can't apply the alternating series estimation theorem to a non-alternating series.

    - Hollywood
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    Re: seriously hitting my boiling point on these problems please help

    Quote Originally Posted by hollywood View Post
    \sum_{n=1}^\infty \frac{(-1)^{n-1}n}{n^4+1} is an alternating series - whose terms decrease in absolute value. But you are correct - you can't apply the alternating series estimation theorem to a non-alternating series.

    - Hollywood
    I'm talking about the sum: sin^2(n)/n^5.
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  8. #8
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    Re: seriously hitting my boiling point on these problems please help

    Yeah - that one's not an alternating series (in fact, all terms are positive), so you need to use johng's method.

    - Hollywood
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    Re: seriously hitting my boiling point on these problems please help

    Quote Originally Posted by johng View Post
    Hi,
    In your statement of an integral test, who is function f? Answer: f is a decreasing non-negative function on say $[1,\infty)$ with $f(n)=a_n$ and $\int_1^\infty f(x)\,dx$ is finite. Here also $R_n=\sum_{k=n+1}^\infty a_n$.

    Your problem:
    $$b_n={sin^2(n)\over n^5}\text{. Then }R_n=|\sum_{k=n+1}^\infty b_n|\leq\sum_{k=n+1}^\infty {1\over k^5}$$

    So let $f(x)={1\over x^5}$. Use this to find an error bound for $\sum_{k=n+1}^\infty {1\over k^5}$ and hence for $R_n$

    Your second series is an alternating series with the terms decreasing.
    I'm sorry but im still very confused by this solution could you please walk me through what ideas you used it seems like a mix of the alternating series therom and comparison test but i cant be sure
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    Re: seriously hitting my boiling point on these problems please help

    He's bounding the error, which is exactly \left| \sum_{k=n+1}^\infty b_n \right| by the easier-to-analyze series \sum_{k=n+1}^\infty \frac{1}{k^5}. That's similar to the comparison test; he's looking at the actual sum instead of just testing for convergence.

    The next step is like the integral test, but again he's looking for the actual sum instead of just testing for convergence. Since the terms are decreasing,

    \sum_{k=n+1}^\infty \frac{1}{k^5} \le \int_n^\infty \frac{1}{x^5}\,dx.

    - Hollywood
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