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Find sum of the series

  1. Jul 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the power series representation of f(x)= 1/((4+x)^2).
    Use this representation to determine the sum of the series:
    sum [1, infinity) of n*(3/8)^n.
    I attached a screenshot of the problem for just in case.

    2. Relevant equations



    3. The attempt at a solution
    What I did first was find the power series representation of f(x)= 1/(4+x).
    I got sum [0, infinity) of ((-1)^n))(x^n)/(4^(n+1)).
    and then I got the derivative of that which made it:
    sum [1, infinity) of ((-1)^n))(n)(x^(n-1))/(4^(n+1)).
    And so that is where I am lost. I don't know how to use that representation to find the sum of the series "n*(3/8)^n".
     

    Attached Files:

  2. jcsd
  3. Jul 2, 2012 #2

    Curious3141

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    Your first error is that the derivative of [itex]\frac{1}{4+x}[/itex] is in fact [itex]-\frac{1}{{(4+x)}^2}[/itex]. You're missing a negative sign in your series (which can then be incorporated into the power of -1 coefficient).

    Another hint: [itex]\frac{x^{n-1}}{4^{n+1}} = \frac{1}{16}\frac{x^{n-1}}{4^{n-1}}[/itex]. Try to put [itex]k = n-1[/itex] and see if you can finesse that series into a simpler form.

    Now try to figure out what value of x would allow you to express that series as sums of powers of [itex]\frac{3}{8}[/itex]. You may need to do a little additional finessing, like recognising that your power series is actually the sum you want plus a geometric series.
     
    Last edited: Jul 2, 2012
  4. Jul 2, 2012 #3
    thank you so much for your help! I have a final today and this type of problem confuses me so much. I couldn't figure it out, but I at least got a little closer to the answer. I made the correction you pointed out about the negative sign. Then I was able to use your hint to get the series to (1/16)(n)((x/4)^(n-1)), but I don't know how to take it any further. Also, I dont know if this is correct but I tried figuring out what the value of x would have to be to get 3/8 and I came up with 3/2.
     
  5. Jul 2, 2012 #4

    Curious3141

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    I'm going to start by expressing the power series in terms of powers of k rather than n, so that the final answer is neater. The choice of letter doesn't matter.

    You seem to have got rid of the (-1)k coefficient entirely. This is not correct. Remember that an alternating series (alternate pos and neg terms), when multiplied by -1 will yield another alternating series, only switching which terms are positive and negative. When you multiply [itex](-1)^k[/itex] by -1, you can express the result as either [itex](-1)^{k+1}[/itex] or [itex](-1)^{k-1}[/itex], your choice, since they are both equal. In this case, it makes sense to choose the latter because it makes simplification by the algebraic substitution n = k-1 easier.

    So the series is [itex]\frac{1}{16}\sum_{k=1}^{\infty}{(-1)}^{k-1}.k.{(\frac{x}{4})}^{k-1}[/itex], and this series sums to [itex]\frac{1}{{(4+x)}^2}[/itex].

    Now substitute n = k - 1. Since k starts from 1, n starts from p. They both end at ∞.

    The series becomes [itex]\frac{1}{16}\sum_{n=0}^{\infty}{(-1)}^{n}.(n+1){(\frac{x}{4})}^{n}[/itex], and this series also sums to [itex]\frac{1}{{(4+x)}^2}[/itex].

    Observe that [itex]\frac{1}{16}\sum_{n=0}^{\infty}{(-1)}^{n}.(n+1){(\frac{x}{4})}^{n} = \frac{1}{16}(1 + \sum_{n=1}^{\infty}{(-1)}^{n}.(n+1){(\frac{x}{4})}^{n})[/itex] (bringing out the first term from the summation), which then becomes [itex]\frac{1}{16}[1 + \sum_{n=1}^{\infty}{(-1)}^{n}.n.{(\frac{x}{4})}^{n} + \sum_{n=1}^{\infty}{(-1)}^{n}{(\frac{x}{4})}^{n}][/itex] (splitting the (n+1) term up).

    The sum [itex]\sum_{n=1}^{\infty}{(-1)}^{n}{(\frac{x}{4})}^{n}[/itex] is actually a geometric series with first term and common ratio both equal to [itex]-\frac{x}{4}[/itex]. So its sum is:

    [tex]\frac{-\frac{x}{4}}{1-(-\frac{x}{4})} = -\frac{x}{4+x}[/tex]

    Hence this equation holds true:

    [tex]\frac{1}{16}[1 + \sum_{n=1}^{\infty}{(-1)}^{n}.n.{(\frac{x}{4})}^{n} -\frac{x}{4+x}] = \frac{1}{{(4+x)}^2}[/tex]

    for the joint radius of convergence of the power series and the geometric series. That of the power series is [itex]|\frac{x}{4}| < 1 \Rightarrow -4 < x < 4 [/itex], which coincides (in this case) with the radius of convergence of the geometric series. So we're all good.

    Here's the slightly tricky bit. We have an alternating series, but the question is asking us to sum a series of positive terms. So we need to cleverly choose x to be [itex]-\frac{3}{2}[/itex] (note the minus sign), so that it multiplies out with the [itex]{(-1)}^n[/itex] to give all positive terms. We trivially verify that this value of x is within the radius of convergence. So put that into the equation:

    [tex]\frac{1}{16}[1 + \sum_{n=1}^{\infty}n.{(\frac{3}{8})}^{n} + \frac{3}{5}] = \frac{4}{25}[/tex]

    You should get that the required sum is [itex]\frac{24}{25}[/itex].

    (which I verified with Wolfram Alpha!) :tongue:

    As an afternote, I'll state for the record that whoever set this question is a friggin' sadist. He/she could easily have started with the power series for [itex]\frac{1}{{(4-x)}^2}[/itex] which would have given a series of positive terms, and allowed a simpler choice of x ([itex]\frac{3}{2}[/itex], as you stated). Instead, the question stuck us with an unnecessarily complicated alternating series as the starting point. :rolleyes:

    I hope your final went fine, and sorry I couldn't help earlier. I was knocked out the whole day with a very bad cold - I'm still woozy and half-asleep now, in fact. :zzz:
     
    Last edited: Jul 2, 2012
  6. Jul 2, 2012 #5
    Thank you, again! I actually got a chance to read your post before my exam, and was able to do the question that was similar to this one on it.
     
  7. Jul 2, 2012 #6

    Curious3141

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    I'm glad. :smile:
     
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