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Explain why a geometric series with a ratio between zero and one converges and how you find the sum.

Verified Answer :

Let \(S_n\) be the \(n\)th partial sum of a geometric sequence with common ratio \(r\) and first term \(a\). So the sequence is \(\{a,ar,ar^2,\cdots\}\), and \(S_n\) is

\(S_n=a+ar+ar^2+\cdots+ar^{n-2}+ar^{n-1}\)

Multiplying both sides by \(r\) gives

\(rS_n=ar+ar^2+ar^3+\cdots+ar^{n-1}+ar^n\)

and subtracting \(rS_n\) from \(S_n\) gives

\(S_n-rS_n=a+(ar-ar)+(ar^2-ar^2)+\cdots+(ar^{n-1}-ar^{n-1})-ar^n\)
\((1-r)S_n=a(1-r^n)\)
\(S_n=a\dfrac{1-r^n}{1-r}\)

If \(0, then \(r^n\to0\) as \(n\to\infty\) and so the sum approaches

\(\displaystyle\lim_{n\to\infty}S_n=\dfrac a{1-r}\)

Answer:

The formula for the sum is Sn= a1(1/r^n)/(1-r)

A fraction raised to a large power approaches zero.

The sum is a1 divided by the difference of 1 and r.