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Thread: Finding convergence of a series???

  1. #1
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    Question Finding convergence of a series???

    ive run into the following question and i really don't know where to start. i havnt come across any questions involving 2 variables yet.

    For what values of x does the following series converge?

    k=1(sigma)infinity = x^k/((4^k)*k)

    any help would be great.
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  2. #2
    Super Member Deadstar's Avatar
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    What you want to do is treat x as a constant and do the ratio test.

    So for $\displaystyle a_k = \frac{x^k}{4^k k}$ we get

    $\displaystyle |\frac{a_{k+1}}{a_k}| = |\frac{x^{k+1}}{4^{k+1} (k+1)} \cdot \frac{4^k k}{x^k}|$ = $\displaystyle |\frac{k}{k+1} \cdot \frac{1}{4} \cdot x|$.

    Now for this to converge you want $\displaystyle |\frac{k}{k+1} \cdot \frac{1}{4} \cdot x| <1$ when k goes to infinity. So when k does go to infinity $\displaystyle \frac{k}{k+1} -> 1$ so we are left with $\displaystyle |\frac{1}{4} \cdot x| <1$
    hence $\displaystyle -4<x<4$.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Jonny-123 View Post
    ive run into the following question and i really don't know where to start. i havnt come across any questions involving 2 variables yet.

    For what values of x does the following series converge?

    k=1(sigma)infinity = x^k/((4^k)*k)

    any help would be great.
    First improve your notation, more readable would be:

    sum( x^k/(k 4^k) , k=1 .. infty)

    better yet learn to use the math type setting facility on this site, see the tutorial here

    The general term of your series is:

    $\displaystyle a_k=\frac{(x/4)^k}{k}$

    when $\displaystyle |x|<4$ this is bounded in absolute value by $\displaystyle (|x|/4)^k$ which is the general term of a convergent geometric series and hence your series is convergent.

    when $\displaystyle x=4$ the series is the harmonic series and hence divergent, when $\displaystyle x=-4$ the series is the alternating harmonic series and hence convergent.

    When $\displaystyle |x|>4$, $\displaystyle a_k$ does not go to zero as $\displaystyle k$ goes to infinity hence the series diverges.

    CB
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