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I need help with understanding how to solve this task, because I'm a bit lost at the moment.

Use the powerseries $$f(x)=\frac{1}{1-x}$$ to decide the sum of the series

$\sum_{n=1}^{\infty} n(n+1)x^n$ and $\sum_{n=1}^{\infty} \frac{n(n+1)}{3^n}$

I don't understand how to manipulate the sums to use the power series of the function.

Mathomat55
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  • Hmm, I don't understand the "logic" when solving the task. So I didn't understand so much from that previous question. Should I not find a number and not only if the sum converge or diverge? – Mathomat55 Feb 25 '20 at 12:11
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    The answer proves that$$\sum_{n=1}^\infty n(n+1)x^n=\frac{2x}{(1-x)^2}.$$What else do you want? – José Carlos Santos Feb 25 '20 at 12:14
  • Okay, I will read through again and see if I understand it ! – Mathomat55 Feb 25 '20 at 12:19
  • Okay, I have tried using the reasoning from link you sent me and tried to differentiate:

    $\frac{1}{1-x}=\sum_{n=0}^{\infty} x^n$

    But when I differentiate twice. I get

    $\frac{2}{(1-x)^3}=\sum_{n=0}^{\infty} n(n-1)x^{n-2}$

    So I don't get the answer in the link. Is this due to the summation index?

     – Mathomat55 Feb 25 '20 at 13:16
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    No, you don't get $\sum_{n=0}^\infty n(n-1)x^{n-2}$. What you get is $\sum_{n=2}^\infty n(n-1)x^{n-2}$, which is equal to $\sum_{n=0}^\infty(n+2)(n+1)x^n$. – José Carlos Santos Feb 25 '20 at 13:40
  • Okay, I think I got it now. I'm sorry by the way. I'm a slow learner, so it takes some time for me to understand. But I really appreciate your help! But what do you think about what I have done now?

    So $\frac{2}{(1-x)^3}=\sum_{n=0}^{\infty} n(n-1)x^{n-2}$ and

    $\frac{2}{(1-x)^3}=\sum_{n=1}^{\infty} n(n+1)x^{n}x^{-1}$

    And then I multiply by x on each side and get

    $\frac{2x}{(1-x)^3}=\sum_{n=1}^{\infty} n(n+1)x^{n}$

     – Mathomat55 Feb 25 '20 at 13:52
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    That is correct indeed. – José Carlos Santos Feb 25 '20 at 14:03
  • Thank you again for your help :) – Mathomat55 Feb 25 '20 at 14:11

1 Answers1

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First, note that for $|x| < 1$ we have $f(x) = \sum_{n=0}^\infty x^n$.

and for each $|r|<1$ the series converge uniformly in $[-r,r]$ so we can use "derivative term by term" to get:

$f'(x) = \sum_{n=0}^\infty nx^{n-1}$. Use this reasoning again to get

$f''(x) = \sum_{n=0}^\infty (n-1)nx^{n-2}$

Use this to calculate $\sum_n n(n+1)3^{-n}$

infinity
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