Results 1 to 5 of 5

Thread: Problem 30

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10

    Problem 30

    Show that $\displaystyle \sum_{n=1}^{\infty}nx^{n-1} = \frac{1}{(1-x)^2}$ for $\displaystyle |x|<1$ without differenciating the geometric power series.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    Let $\displaystyle S_n=\sum_{k=1}^nkx^{k-1}$
    We have $\displaystyle S_n=1+2x+3x^2+\ldots +nx^{n-1}$ (1).
    Multiplying (1) with $\displaystyle x$ we have
    $\displaystyle xS_n=x+2x^2+3x^3+\ldots +(n-1)x^{n-1}+nx^n$ (2)
    Substracting (2) from (1) yields
    $\displaystyle S_n(1-x)=1+x+x^2+\ldots +x^{n-1}-nx^n=\frac{1-x^n}{1-x}-nx^n\Rightarrow S_n=\frac{1-x^n}{(1-x)^2}-\frac{nx^n}{1-x}$
    Then $\displaystyle \sum_{n=1}^{\infty}nx^{n-1}=\lim_{n\to\infty}S_n=\frac{1}{(1-x)^2}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Since you answered so quickly, here is a second challenge. I am not sure if is a fair question to ask but I post it anyway.
    ---

    Having shown that $\displaystyle \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}$

    Can you prove the following?
    Let $\displaystyle f(x) = \sum_{n=0}^{\infty} a_nx^n$ with radius of convergence $\displaystyle R>0$

    Then, $\displaystyle f$ is differenciable on $\displaystyle (-R,R)$ and furthermore $\displaystyle f'(x) = \sum_{n=1}^{\infty} na_nx^{n-1}$.

    Note: This works even if $\displaystyle x$ is a complex number, but the proofs are completely anagolous.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    I am not sure if is a fair question to ask but I post it anyway.
    That's ok, we just apply the relevant theorem
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Rebesques View Post
    That's ok, we just apply the relevant theorem
    I am not sure what you mean. In my Analysis class we proved differenciation term-by-term by working backwards with the Fundamental Theorem of Calculus after proving integration term-by-term. And that was just Real fucntions.


    By my Complex Analysis book had a really nice approach to this proof. Without using more advandeced techiqnues. It was able to prove term-by-term differenciation by using the series posted above. It was a long but straightforward derivation. That is what I am asking to show.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum