Sunday, 15 September 2013

Sum of a Geometric series and its Application

Sum

The sum of a geometric series is finite as long as the terms approach zero; as the numbers near zero, they become insignificantly small, allowing a sum to be calculated despite the series being infinite. The sum can be computed using the self-similarity of the series.

Example



Consider the sum of the following geometric series:
s \;=\; 1 \,+\, \frac{2}{3} \,+\, \frac{4}{9} \,+\, \frac{8}{27} \,+\, \cdots
This series has common ratio 2/3. If we multiply through by this common ratio, then the initial 1 becomes a 2/3, the 2/3 becomes a 4/9, and so on:
\frac{2}{3}s \;=\; \frac{2}{3} \,+\, \frac{4}{9} \,+\, \frac{8}{27} \,+\, \frac{16}{81} \,+\, \cdots
This new series is the same as the original, except that the first term is missing. Subtracting the new series (2/3)s from the original series s cancels every term in the original but the first:
s \,-\, \frac{2}{3}s \;=\; 1,\;\;\;\mbox{so }s=3.
A similar technique can be used to evaluate any self-similar expression.

Formula

For r\neq 1, the sum of the first n terms of a geometric series is:
a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r},
where a is the first term of the series, and r is the common ratio. We can derive this formula as follows:

\begin{align}
&\text{Let }s = a + ar + ar^2 + ar^3 + \cdots + ar^{n-1}. \\[4pt]
&\text{Then }rs = ar + ar^2 + ar^3 + ar^4 + \cdots + ar^{n}  \\[4pt]
&\text{Then }s - rs = a-ar^{n}  \\[4pt]
&\text{Then }s(1-r) = a(1-r^{n}),\text{ so }s = a \frac{1-r^{n}}{1-r} \quad \text{(if } r \neq 1 \text{)}.
\end{align}
As n goes to infinity, the absolute value of r must be less than one for the series to converge. The sum then becomes
a+ar+ar^2+ar^3+ar^4+\cdots = \sum_{k=0}^\infty ar^k = \frac{a}{1-r} \Leftrightarrow |r|<1
When a = 1, this simplifies to:
1 \,+\, r \,+\, r^2 \,+\, r^3 \,+\, \cdots \;=\; \frac{1}{1-r},
the left-hand side being a geometric series with common ratio r. We can derive this formula:

\begin{align}
&\text{Let }s = 1 + r + r^2 + r^3 + \cdots. \\[4pt]
&\text{Then }rs = r + r^2 + r^3 + \cdots. \\[4pt]
&\text{Then }s - rs = 1,\text{ so }s(1 - r) = 1,\text{ and thus }s = \frac{1}{1-r}.
\end{align}
The general formula follows if we multiply through by a.
This formula is only valid for convergent series (i.e., when the magnitude of r is less than one). For example, the sum is undefined when r = 10, even though the formula gives s = −1/9.
This reasoning is also valid, with the same restrictions, for the complex case.

Proof of convergence

We can prove that the geometric series converges using the sum formula for a geometric progression:
\begin{align}
&1 \,+\, r \,+\, r^2 \,+\, r^3 \,+\, \cdots \\[3pt]
&=\; \lim_{n\rightarrow\infty} \left(1 \,+\, r \,+\, r^2 \,+\, \cdots \,+\, r^n\right) \\
&=\; \lim_{n\rightarrow\infty} \frac{1-r^{n+1}}{1-r}
\end{align}
Since (1 + r + r2 + ... + rn)(1−r) = 1−rn+1 and rn+1 → 0 for | r | < 1.
Convergence of geometric series can also be demonstrated by rewriting the series as an equivalent telescoping series. Consider the function g(K) = (r^(K+1))/(1-r). Note that: r = g(0) - g(1), r^2 = g(1) - g(2), r^3 = g(2) - g(3), . . . Thus: S = r + r^2 + r^3 + . . . = (g(0) - g(1)) + (g(1) - g(2)) + (g(2) - g(3)) + . . . If |r|<1, then g(K) -> 0 as K -> infinity, and so S converges to g(0) = r/(1-r).

Applications

Repeating decimals

A repeating decimal can be thought of as a geometric series whose common ratio is a power of 1/10. For example:
0.7777\ldots \;=\; \frac{7}{10} \,+\, \frac{7}{100} \,+\, \frac{7}{1000} \,+\, \frac{7}{10000} \,+\, \cdots.
The formula for the sum of a geometric series can be used to convert the decimal to a fraction:
0.7777\ldots \;=\; \frac{a}{1-r} \;=\; \frac{7/10}{1-1/10} \;=\; \frac{7}{9}.
The formula works not only for a single repeating figure, but also for a repeating group of figures. For example:
0.123412341234\ldots \;=\; \frac{a}{1-r} \;=\; \frac{1234/10000}{1-1/10000} \;=\; \frac{1234}{9999}.
Note that every series of repeating consecutive decimals can be conveniently simplified with the following:
0.09090909\ldots \;=\; \frac{09}{99} \;=\; \frac{1}{11}.
0.143814381438\ldots \;=\; \frac{1438}{9999}.
0.9999\ldots \;=\; \frac{9}{9} \;=\; 1.
That is, a repeating decimal with repeat length n is equal to the quotient of the repeating part (as an integer) and 10n - 1.

To Join Ajit Mishra's Online Classroom  CLICK HERE

No comments:

Post a Comment

Our Latest Post

How to Evaluate an Integral

In this video you will see how to evaluate an Integral. This video shows an example, by this example you will see about how to evaluate an ...

Popular Post