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Thread: Geometric Series - Converge or Diverge?

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    Geometric Series - Converge or Diverge?

    $\displaystyle \sum_{n = 0}^{\infty}\dfrac{3}{10^{n + 1}}$

    $\displaystyle r = \dfrac{1}{10}$ but how can we find r? It seems like somebody said to divide one term in the series by the one before. But doing so doesn't come out to $\displaystyle \dfrac{1}{10}$

    Here is the formula for $\displaystyle r$:

    $\displaystyle r = \dfrac{a_{n+1}}{a_{n}}$ What would go in the numerator and denominator, in this case?

    Anyhow since $\displaystyle \dfrac{1}{10} < 1$ then the geometric series does NOT diverge

    But here is another question. Now, we have to find the sum. Is it possible that, after this point, the series could diverge?

    $\displaystyle S = \dfrac{a}{1 - r}$

    $\displaystyle S = \dfrac{\dfrac{3}{10}}{1 - \dfrac{1}{10}} = \dfrac{1}{3}$ - The infinite series converges. But what does this have to do with the sum?
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    Re: Geometric Series - Converge or Diverge?

    Quote Originally Posted by Jason76 View Post
    $\displaystyle \sum_{n = 0}^{\infty}\dfrac{3}{10^{n + 1}}$

    $\displaystyle r = \dfrac{1}{10}$ but how can we find r? It seems like somebody said to divide one term in the series by the one before. But doing so doesn't come out to $\displaystyle \dfrac{1}{10}$
    Yes, it does. "One term in the series", taking n= k is $\displaystyle \dfrac{3}{10^{k+1}}$. The "one before" is $\displaystyle \dfrac{3}{10^{(k-1)+ 1}}= \dfrac{3}{10^k}$. So "one term in the series, divided by the one before" is $\displaystyle \dfrac{3}{10^{k+1}}\dfrac{10^k}{3}= \dfrac{1}{10}$.

    Here is the formula for $\displaystyle r$:

    $\displaystyle r = \dfrac{a_{n+1}}{a_{n}}$ What would go in the numerator and denominator, in this case?

    Anyhow since $\displaystyle \dfrac{1}{10} < 1$ then the geometric series does NOT diverge

    But here is another question. Now, we have to find the sum. Is it possible that, after this point, the series could diverge?

    $\displaystyle S = \dfrac{a}{1 - r}$

    $\displaystyle S = \dfrac{\dfrac{3}{10}}{1 - \dfrac{1}{10}} = \dfrac{1}{3}$ - The infinite series converges. But what does this have to do with the sum?
    The same thing "2" has to do with "1+ 1"! It is the value of the sum: $\displaystyle \sum_{n=0}^\infty \dfrac{3}{10^{n+1}}= 3$.
    Last edited by HallsofIvy; Dec 28th 2013 at 04:35 PM.
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    Re: Geometric Series - Converge or Diverge?

    Quote Originally Posted by HallsofIvy View Post
    It is the value of the sum: $\displaystyle \sum_{n=0}^\infty \dfrac{3}{10^{n+1}}= 3$.
    I think you meant to say $\displaystyle \sum_{n=0}^\infty \dfrac{3}{10^{n+1}}=\frac{1}{3}$, right?

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    Re: Geometric Series - Converge or Diverge?

    Given $\displaystyle n = 0$ That is, 0 is the first term in the series.

    $\displaystyle a_{n} = a_{0} = \dfrac{3}{10^{(0) + 1}} = \dfrac{3}{10^{1}} = \dfrac{3}{10}$

    $\displaystyle a_{n + 1} = a_{1} = \dfrac{3}{10^{(1) + 1}} = \dfrac{3}{10^{2}} = \dfrac{3}{100}$

    $\displaystyle \dfrac{a_{n + 1}}{a_{n}} = \dfrac{a_{1}}{a_{0}} = \dfrac{\dfrac{3}{100}}{ \dfrac{3}{10}} = \dfrac{1}{10}$
    Last edited by Jason76; Dec 28th 2013 at 10:24 PM.
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    Re: Geometric Series - Converge or Diverge?

    Quote Originally Posted by Jason76 View Post
    Given $\displaystyle n = 0$ That is, 0 is the first term in the series.

    $\displaystyle a_{n} = a_{0} = \dfrac{3}{10^{(0) + 1}} = \dfrac{3}{10^{1}} = \dfrac{3}{10}$

    $\displaystyle a_{n + 1} = a_{1} = \dfrac{3}{10^{(1) + 1}} = \dfrac{3}{10^{2}} = \dfrac{3}{100}$

    $\displaystyle \dfrac{a_{n + 1}}{a_{n}} = \dfrac{a_{1}}{a_{0}} = \dfrac{\dfrac{3}{100}}{ \dfrac{3}{10}} = \dfrac{1}{10}$
    Yes, so $\displaystyle \frac{a_{1}}{a_{0}} = \frac{1}{10}$. But what HallsofIvy is saying is much more. He says that $\displaystyle \frac{a_{n+1}}{a_{n}} = \frac{1}{10}$ for all possible values of $\displaystyle n$:

    $\displaystyle \frac{a_{n+1}}{a_{n}} = \frac{\frac{3}{10^{n+1}}}{\frac{3}{10^n}} = \frac{3}{10^{n+1}}\frac{10^n}{3} = \frac{3}{(10)(10^n)}\frac{10^n}{3} = \frac{1}{10}$

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