Demonstrate the expression is valid for \(n=1\) ( i.e. \(u_1\) = 2). \[ u_n=3^n-1 \] \[ u_1=3^1-1 \] \[ u_1=3 -1 = 2 \]
Demonstrate the expression is valid for \(n=k\) . \[ u_k=3^k-1 \]
Demonstrate the expression is valid for \(n=k+1\) .
\[ u_{k+1}=3^{k+1}-1 \] \[ u_{k+1}=3.3^{k}-1 \] \[ u_{k+1}=(1+2).3^{k}-1 \] \[ u_{k+1}=3^{k} + 2.3^{k}-1 \]
An infinite series is a series whose terms are summed indefinitely. For an infinite sequence \(a_0,a_1,a_2,a_3,\ldots,a_n,\ldots\), the corresponding infinite series \(s_\infty\) is written \[ s_\infty=\sum_{n=0}^{\infty}a_n=a_0+a_1+a_2+a_3+a_4+a_5+\cdots \] Formally, an infinite series can be expressed as the limit of the series as \(n\to \infty\) \[ s_\infty = \lim_{n \to \infty} s_n = \lim_{n \to \infty} \sum_{k=0}^n a_k \] For any particular series, this limit may or may not exist. If it does exist, then \(s_\infty\) is defined and the infinite series is . If the limit does not exist then \(s_\infty\) is not defined and the infinite series is divergent.
The infinite series \[ s_\infty = 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots = 2 \] Letting \(a_n=\left(\frac{1}{2}\right)^n\), \(s_\infty\) is the limit of the sequence \(s_n=\sum_{k=0}^{n} \left(\frac{1}{2}\right)^k\). Applying the formula for the sum of a geometric series gives
\[ \begin{align*} s_n = \sum_{k=0}^{n} \frac{1}{2^k} = \frac{1- \left( \frac{1}{2} \right)^{n+1} }{1-\frac{1}{2}} = 2 \left[ 1-\frac{1}{2^{n+1}} \right] \end{align*} \]
And so applying the definition of the infinite series gives. \[ s_\infty=\lim_{n \to \infty} s_n = \lim_{n \to \infty} 2 \left[ 1- \frac{1}{2^{n+1}} \right]= 2 \left[ 1-0 \right] = 2 \]\ A condition for the convergence of an infinite series is that the terms in the sequence \(a_n \to 0\) as \(n \to \infty\). However, this condition is not to ensure convergence.
Consider the harmonic sequence \(a_n=1/n\) and the corresponding infinite harmonic series \(\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}\). This series is divergent. To see this compare the following \(2\) infinite series
\[\begin{align*} &1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\cdots+\frac{1}{16}+\frac{1}{17}\cdots\\ >&1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\cdots+\frac{1}{16}+\frac{1}{32}\cdots\\ = & 1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \end{align*} \]
The first series is the harmonic series and the terms in the second series are less than or equal to the corresponding terms in the harmonic series. However, the second series diverges as it is the sum of an infinite number of \(1/2\) terms. So the harmonic series is always greater than a series which diverges, and so the harmonic series itself is divergent.
So even though the terms \(a_n \to 0\), the addition of an infinite number of them can still cause a series to diverge as $n $. A better test for whether a series converges is needed.
To determine whether an infinite series is convergent, the ratio test can be applied. The test examines the ratio of successive terms in the sequence as \(n \to \infty\). If this ratio has a limit, the sequences behaves like a geometric sequence for large \(n\), and so if the ratio is less than \(1\), the series will be convergent.
Formally, given a sequence \(a_n\) and the corresponding infinite series \(\displaystyle \sum_{n=0}^\infty a_n\). Let \[ R=\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \] If this limit \(R\) exists, then
if \(R<1\), the series is convergent. if \(R>1\), the series is divergent. if \(R=1\) or \(R\) does not exist, then the ratio test was inconclusive.
Let \(a_n=\dfrac{1}{n!}\), and consider the infinite series \(\displaystyle \sum_{n=0}^{\infty} \frac{1}{n!} = 1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots\). Since \(a_{n+1}=\dfrac{1}{(n+1)!}\) \[ \left| \frac{a_{n+1}}{a_n}\right|=\left| \frac{1}{(n+1)!}\cdot \frac{n!}{1}\right|=\left| \frac{n!}{(n+1)!}\right|=\left| \frac{1}{n+1}\right|=\frac{1}{n+1} \] So to evaluate \(R\), \[ R=\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty} \frac{1}{n+1}=0 \] So \(R=0\) meaning that \(R<1\), and therefore the infinite series $ _{n=0}^{} $ is convergent.
Let \(a_n=\dfrac{2^n}{n^2}\), and consider the infinite series \(\displaystyle \sum_{n=0}^{\infty} \frac{2^n}{n^2}\). Since \(a_{n+1}=\dfrac{2^{n+1}}{(n+1)^2}\) \[ \left| \frac{a_{n+1}}{a_n}\right|=\left| \frac{2^{n+1}}{(n+1)^2}\cdot \frac{n^2}{2^n}\right|=\left| \frac{2^{n+1}}{2^n}\cdot\frac{n^2}{(n+1)^2}\right|=2 \left(\frac{n}{n+1}\right)^2 \] So to evaluate \(R\), \[ R=\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty} 2 \left(\frac{n}{n+1}\right)^2=\lim_{n \to \infty} 2 \left(\frac{1}{1+\frac{1}{n}}\right)^2=2\left(\frac{1}{1+0}\right)^2=2 \] So \(R=2\) meaning \(R>1\), and therefore the infinite series \(\displaystyle \sum_{n=0}^{\infty} \frac{2^n}{n^2}\) is divergent.
Convergent infinite series are often used to define functions. The \(y=e^x=\text{exp}(x)\) is defined as \[ \text{exp}(x) = \sum_{n=0}^{\infty} \frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\]
The ratio test can be used to show that this infinite series is convergent for all values of \(x\). Here, the sequence terms are \(a_n=\dfrac{x^n}{n!}\), so \(a_{n+1}=\dfrac{x^{n+1}}{(n+1)!}\), and \[ \left| \frac{a_{n+1}}{a_n}\right|=\left| \frac{x^{n+1}}{(n+1)!}\cdot \frac{n!}{x^n}\right|=\left| \frac{x^{n+1}}{x^n}\cdot\frac{n!}{(n+1)!}\right|=\left|x \frac{1}{n+1}\right| = \frac{|x|}{n+1} \] So to evaluate \(R\), \[ R=\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|=\lim_{n \to \infty} \frac{|x|}{n+1}=0 \] So \(R<1\) and the series is convergent for all \(x\).
The infinite series can be used to approximate \(e^x\) to as many decimal places desired. To do this, use the sequence terms to evaluate the terms in the series \(\displaystyle s_n=\sum_{k=0}^{n} \frac{x^k}{k!}\). This is best done using a table similar to the one below.
For example, to evaluate \(e^{0.3}\) to three decimal places, the terms in the series \(\displaystyle s_n=\sum_{k=0}^{n} \frac{(0.3)^k}{k!}\) must be computed until the desired accuracy is reached. %\end{center} The series can be summed indefinitely but in practice some kind of stopping condition is needed. A heuristic stopping condition can be used; for example, we can stop when the terms \(a_n\) become small enough, or when the digits of \(s_n\) appear to have converged to enough decimal places. The first condition assumes that the terms \(a_n\) never become large again, and the second condition may have problems if the terms \(a_n\) alternate around \(0\). However, in practice, we can stop if the series appears to have converged after a few iterations.
In this case, to \(3\) decimal places \(e^{0.3} \cong 1.349\). A more accurate estimate would be \(e^{0.3}\cong 1.34985880757600310398\).
In mathematics, the ratio test is a test (or “criterion”) for the convergence of a series \[ \sum_{n=1}^\infty a_n,\] where each term is a real or complex number and \(a_n\) is nonzero when n is large.
The test was first published by Jean le Rond d’Alembert and is sometimes known as d’Alembert’s ratio test or as the Cauchy ratio test.
Given the following geometric series: \[ \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\]
The quotient $ a_{n+1}/a_n = (1/2){n+1}/(1/2)n$ of any two adjacent terms is 1/2. The sum of the first m terms is given by: \[1 - \frac{1}{2^m}.\]
As m increases, this converges to 1, so the sum of the series is 1. On the other hand given this geometric series: \[\sum_{n=1}^\infty 2^n = 2 + 4 + 8 + \cdots\]
Whether this converges or diverges as m increases depends on whether r, the quotient of any two adjacent terms, is less than or greater than 1. Now consider the series: \[\sum_{n=1}^\infty \frac{n+1}{n} \left(\frac{1}{2}\right)^n = \frac{2}{1} \cdot \frac{1}{2} + \frac{3}{2} \cdot \frac{1}{4} + \frac{4}{3} \cdot \frac{1}{8} + \cdots\]
This is similar to the first convergent sequence above, except that now the ratio of two terms is not fixed at exactly 1/2: \[ \left(\frac{n+1}{n} \left(\frac{1}{2}\right)^n\right)/\left(\frac{n}{n-1} \left(\frac{1}{2}\right)^{n-1}\right) = \frac{n^2-1}{2n^2} = \frac{1}{2} - \frac{1}{2n^2}.\]
The usual form of the test makes use of the limit \[L = \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|.\]
The ratio test states that:
It is possible to make the ratio test applicable to certain cases where the limit L fails to exist, if limit superior and limit inferior are used.
The test criteria can also be refined so that the test is sometimes conclusive even when L = 1. More specifically, let R = ||andr = ||. Then the ratio test states that:
if R < 1, the series converges absolutely; if r > 1, the series diverges; if || for all large n (regardless of the value of r ), the series also diverges; this is because |a_n| is nonzero and increasing and hence a_n does not approach zero; the test is otherwise inconclusive. If the limit L in (1) exists, we must have L=R=r. So the original ratio test is a weaker version of the refined one. ExamplesConvergent because L<1 Consider the series {n=1}^ Putting this into the ratio test: L = {n} | | = _{n} | | = < 1. Thus the series converges.
Divergent because L>1 Consider the series {n=1}^. Putting this into the ratio test: L = {n} | | = _{n} | | = e > 1. Thus the series diverges.
Inconclusive because L=1 Consider the three series {n=1},{n=1}^ and{n=1}(-1)^n. The first series diverges, the second one converges absolutely and the third one converges conditionally. However, the term-by-term magnitude ratios || of the three series are respectively 1,and . So, in all three cases, we have{n}||=1. This illustrates that when L=1, the series may converge or diverge and hence the original ratio test is inconclusive. For the first series _{n=1}^, however, as the term-by-term magnitude ratio ||=1 for all n, we can apply the third criterion in the refined version of the ratio test to conclude that the series diverges.