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    Hi

    I'm utterly stumped by this:

    Find the exact range of x for which the sum to infinity (from n = 1) of (x^n)/( (4^n)n) converges.

    Any pointers would be very much appreciated.

    Thanks in advance
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    (Original post by *Georgine*)
    Hi

    I'm utterly stumped by this:

    Find the exact range of x for which the sum to infinity (from n = 1) of (x^n)/( (4^n)n) converges.

    Any pointers would be very much appreciated.

    Thanks in advance
    If you have s serioes of
    \displaystyle \sum^{\infty}_{n=1}a_n(x-A)^n
    then the radius of convergence (R) around 'A' is
    \frac{1}{R}=lim_{n \to \infty}\sqrt[n]{a_n}
    In your example A=0, At the limit you can use any criteria of convergence,
    for example the ratio of \frac{a_{n+1}}{a_n}
    Note that lim_{n \to \infty}\sqrt[n]{n}=1
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    (Original post by ztibor)
    If you have s serioes of
    \displaystyle \sum^{\infty}_{n=1}a_n(x-A)^n
    then the radius of convergence (R) around 'A' is
    \frac{1}{R}=lim_{n \to \infty}\sqrt[n]{a_n}
    In your example A=0, At the limit you can use any criteria of convergence,
    for example the ratio of \frac{a_{n+1}}{a_n}
    Note that lim_{n \to \infty}\sqrt[n]{n}=1
    I really do not understand this. How does the ratio give me the exact values of x? And what would be the a (little n) in my case?
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    You are considering  \displaystyle \sum_{n=1}^{\infty} a_n , with  a_n = \frac{x^n}{4^n n} .

    By the ratio test:

    - if  \displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 then the series converges.

    - if   \displaystyle \lim_{n \to \infty}   \left| \frac{a_{n+1}}{a_n} \right| > 1 then the series diverges.

    - if  \displaystyle \lim_{n \to \infty}   \left| \frac{a_{n+1}}{a_n} \right| = 1 , then you can't tell anything and need to examine this case separately.

    Use the first bullet point to find what range x has to be in to make the ratio less than 1 - you know that the series converges for all x in this range.

    Then examine the case where the ratio = 1 directly i.e. find the value of x that makes the ratio 1, and see if you think the series converges/diverges for this particular value of x.
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    (Original post by Daniel Freedman)
    Then examine the case where the ratio = 1 directly i.e. find the value of x that makes the ratio 1, and see if you think the series converges/diverges for this particular value of x.
    Given that there were no restrictions on where x lies, I would say 'values of x that make the ratio 1'.
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    (Original post by ljfrugn)
    Given that there were no restrictions on where x lies, I would say 'values of x that make the ratio 1'.
    Indeed. Thanks
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    (Original post by *Georgine*)
    I really do not understand this. How does the ratio give me the exact values of x? And what would be the a (little n) in my case?
    Not the ratio gives but the limit of ratio does a value which maybe finite or infinite or even 0. This value is the reciprotial of R radius,
    For example if the limit is L then
    \displaystyle R=\frac{1}{L}
    and the series will convergent if
    x \in [A-R, A+R]
    For your example:
    \displaystyle \sum^{\infty}_{n=1} \frac{x^n}{4^n \cdot n}
    so
    \displaystyle a_n=\frac{1}{4^n \cdot n}
    With the nth root test
    \displaystyle \frac{1}{R}=lim_{n \to \infty}\sqrt[n]{\frac{1}{4^n \cdot n}}=lim_{n \to \infty}\frac{1}{4 \cdot \sqrt[n]{n}}=\frac{1}{4}
    with the ratio
    \displaystyle \frac{1}{R}=lim_{n \to \infty}\frac{1}{4^{n+1} \cdot (n+1)} \cdot \frac{4^n \cdot n}{1}=lim_{n \to \infty}\frac{4^n \cdot n}{4\cdot 4^n \cdot (n+1)}=
    \displaystyle =lim_{n \to \infty}\frac{1}{4 \cdot (1+\frac{1}{n})}=\frac{1}{4}


    Calculating the limit with R you will get the range of convergence and consider A=0 (convergence around zero)
 
 
 
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