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 January 29th, 2011, 05:26 AM #1 Senior Member   Joined: Oct 2010 Posts: 126 Thanks: 0 Divergence and convergence I'm not sure if I'm heading down the right path here. I'd gladly appreciate some help. $a.)\ \rm{Show that all arithmetic progressions are divergent.}$ $\rm{a_{n}= a + (n - 1)d = (a - d) + nd.}$ $\rm{When d \neq 0, |a_n| \rightarrow \infty\ as n \rightarrow \infty, so the series diverges.}$ $b.)\ \rm{Show that a geometric progression \{a, ar, ar^2, ar^3,...\} is convergent if |r| < 1.}$ $\rm{a_n= a, ar, ar^2, ar^3,...}$ $\rm{Using the squeeze principle: -1 < a_n < 1}$ $\displaystyle \rm{Suppose a_n=< 1 then let a_n= \frac{1 - a_n}{2} > 0....stuck}=$
 January 29th, 2011, 09:44 AM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Divergence and convergence Well, we could just come out and sum one. For a geometric series with ratio r (and I will assume WLOG that the first term is 1), Consider S_n = 1 + r + r^2 + r^3 +...+ r^n Then (S_n)r = r + r^2 + r^3 + ... + r^(n+1). Solve for S, and take a limit.
January 29th, 2011, 04:23 PM   #3
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Quote:
 Originally Posted by DLowry For a geometric series with ratio r (and I will assume WLOG that the first term is 1), Consider S_n = 1 + r + r^2 + r^3 +...+ r^n Then (S_n)r = Solve for S, and take a limit.
$S_{n}r= r + r^2 + r^3 + ... + r^{(n+1)}$
$S_{n}= r^2 + r^3 + ... + r^{(n+1)}$

How does this 'take a limit' concept work?
What exactly is meant by without any loss of generality?

I still don't get part a.

 January 29th, 2011, 10:35 PM #4 Senior Member   Joined: Oct 2010 Posts: 126 Thanks: 0 $b.)\ \rm{Show that a geometric progression \{a, ar, ar^2, ar^3,...\} is convergent if |r| < 1.}$ $\{a,\ ar,\ ar^2,\ ar^3,...\}$ Now $\displaystyle S_n= \frac{a(1 - r^n)}{1 - r}$ and if |r| < 1, then $\displaystyle \lim_{n\to \infty} r^n= 0$ So $\displaystyle\lim_{n\to \infty} S_n= \displaystyle\lim_{n\to \infty} \left[\frac{a(1 - r^n)}{1 - r}\right] = \frac{a}{1 - r}$ If |r| > 1, $\displaystyle \lim_{n\to \infty} r^n= \infty$ and the series does not converge. $\therefore$, provided that |r| < 1, a Geometric Progression converges to a sum of $\frac{a}{1 - r}.$

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