ijairath
1 Answer
Deleted User
A geometric progression sum: 1/1-r (assuming sequence is r, r^2,...r^n
? this is only valid when you take the limit of the sum of a geometric sequence as {n\rightarrow\infty}, and you need to have the first term of the sequence on the numerator, unless it is exactly 1.
that is
\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\in fty}b_1\frac{1-q^{n}}{1-q}=\frac{b_1}{1-q} for |q|<1 and
\lim_{n\rightarrow\infty}S_n=\infty when |q|>1
| 14 years ago. Rating: 1 | |
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