My Math Forum Sequences and Series

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October 29th, 2011, 08:57 AM   #1
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Sequences and Series

The majority of the question is answered and so I only need help with the final part; mathcing my 'sum to infinity' answer to what the text book has. Can anyone help me?

Many thanks.
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 October 29th, 2011, 10:14 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Sequences and Series I wouldn't write $\sum_{n=1}^{n}$; same letter for variable and upperbound. I'll choose k for the variable. You have: a = x and r = qx $\sum_{k=1}^{n} q^{k-1} \cdot x^k = \frac{a(1-r^{n+1})}{1-r} = \frac{x(1-(qx)^{n+1})}{1-qx}$ (note the different outcome; you had $\frac{x(1-(qx)^{n})}{1-qx}$ instead!) Now, you are also given: "q and x are both positive fractions less than one" so 0 < qx < 1 Find $\lim_{n \to \infty}\frac{x(1-(qx)^{n+1})}{1-qx}$ because the "upperbound" becomes $\infty$ as seen in $\sum_{k=1}^{\infty}$ since $\lim_{n \to \infty}(qx)^{n+1}= 0$ because |qx|<1, you have $\lim_{n \to \infty}\frac{x(1-(qx)^{n+1})}{1-qx}= \frac{x(1-0)}{1-qx} = \frac{x}{1-qx}$ Got it?
 October 29th, 2011, 10:22 AM #3 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Sequences and Series Thank you.

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