Sum of a Geometric Sequence Formula Proof

The sum of the first n terms of a geometric sequence is given by:

\[S_n=\frac{a(1-r^n)}{1-r}\]

where the first term is a and the common ratio is r. The sum to infinity is given by:

\[S_\infty = \frac{a}{1-r}\]

given that `abs(r)<1`.

 

Proof for \(S_n\)

Suppose we have a general geometric sequence with first term a and a common ratio r. The sum of the first n terms is given by:

\[S_n=a+ar+ar^2+ \cdots +ar^{n-3}+ar^{n-2}+ar^{n-1}\]

 

Now consider \(rS_n\):

\[rS_n=ar+ar^2+ar^3+ \cdots +ar^{n-2}+ar^{n-1}+ar^n\]

\[S_n-rS_n=[a+ar+ar^2+ \cdots +ar^{n-3}+ar^{n-2}+ar^{n-1}]-[ar+ar^2+ar^3+ \cdots +ar^{n-2}+ar^{n-1}+ar^n]\]

\[S_n-rS_n=a-ar^n\]

\[S_n(1-r)=a-ar^n\]

\[S_n=\frac{a-ar^n}{1-r}\]

\[\therefore S_n=\frac{a(1-r^n)}{1-r}\]

\[Q.E.D.\]

 

Proof for \(S_\infty\)

We know that:

\[S_n=\frac{a(1-r^n)}{1-r}\]

If `abs(r)<1`, as \(n \to \infty\) ,  \(r^n \to 0\) and \((1-r^n) \to 1\) . Therefore:

\[\lim_{n \to \infty} S_n = \frac{a}{1-r}\]

\[\therefore S_\infty = \frac{a}{1-r}\]

\[Q.E.D.\]

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