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When is this super duper geometric series formula valid?
A student in ECE301 once wrote the following formula on his exam:
$ \sum_{n = M}^N \alpha^n = \frac{\alpha^M - \alpha^{N-1}}{(1 - \alpha)} $
Is this formula correct? For what values of the parameters is the formula valid? Please comment.
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Answer 1
First we know the summation of an infinity geometric series: $ \sum_{n=0}^{\infty} \alpha^n = \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq1)
so we can compute
$ \sum_{n=M}^{\infty} \alpha^n = \left( \alpha \right)^M \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq2)
similarly,
$ \sum_{n=N+1}^{\infty} \alpha^n = \left( \alpha \right)^{N+1} \frac{1}{(1 - \alpha)} , \left| \alpha \right| < 1 $; (eq3)
then we can substract eq3 from eq2, if N+1> M
$ \sum_{n=M}^{\infty} \alpha^n - \sum_{n=N+1}^{\infty} \alpha^n = \frac{{\left( \alpha \right)^M } - {\left( \alpha \right)^{N+1}}}{(1 - \alpha)} , \left| \alpha \right| < 1 $;
for N larger or equal to M, $ \left| \alpha\right| < 1 $, the equation above holds.
//Did I make any mistake in the N+1 part?
//ahhh, the series is finite, so the condition $ \left| \alpha\right| < 1 $ doesn't necessarily hold. but it requires an other approach with
$ \sum_{n=-\infty}^{0} \alpha^n = \frac{1}{(1 - \frac{1}{\alpha})} , \left| \alpha \right| > 1 $;
Anyone want to finish it for me? Thanks
Answer 2
Write it here.
Answer 3
Write it here
