# Thread: Geometric Series - Converge or Diverge?

1. ## Geometric Series - Converge or Diverge?

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

$r = \dfrac{1}{10}$ 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 $\dfrac{1}{10}$

Here is the formula for $r$:

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

Anyhow since $\dfrac{1}{10} < 1$ 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?

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

$S = \dfrac{\dfrac{3}{10}}{1 - \dfrac{1}{10}} = \dfrac{1}{3}$ - The infinite series converges. But what does this have to do with the sum?

2. ## Re: Geometric Series - Converge or Diverge?

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

$r = \dfrac{1}{10}$ 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 $\dfrac{1}{10}$
Yes, it does. "One term in the series", taking n= k is $\dfrac{3}{10^{k+1}}$. The "one before" is $\dfrac{3}{10^{(k-1)+ 1}}= \dfrac{3}{10^k}$. So "one term in the series, divided by the one before" is $\dfrac{3}{10^{k+1}}\dfrac{10^k}{3}= \dfrac{1}{10}$.

Here is the formula for $r$:

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

Anyhow since $\dfrac{1}{10} < 1$ 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?

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

$S = \dfrac{\dfrac{3}{10}}{1 - \dfrac{1}{10}} = \dfrac{1}{3}$ - 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: $\sum_{n=0}^\infty \dfrac{3}{10^{n+1}}= 3$.

3. ## Re: Geometric Series - Converge or Diverge?

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

- Hollywood

4. ## Re: Geometric Series - Converge or Diverge?

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

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

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

$\dfrac{a_{n + 1}}{a_{n}} = \dfrac{a_{1}}{a_{0}} = \dfrac{\dfrac{3}{100}}{ \dfrac{3}{10}} = \dfrac{1}{10}$

5. ## Re: Geometric Series - Converge or Diverge?

Originally Posted by Jason76
Given $n = 0$ That is, 0 is the first term in the series.

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

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

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

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

- Hollywood