Ratio Test For Convergence
A sequence is a function from the set of natural numbers NN to the set of real numbers RR
. That is any function f : NN - RR is called a sequence. For each n in NN , f(n) is well defined. We re-denote f(n) := xn and write the range of f as { xn } .Its simply a new notation, for us xn simply means that it is the value of f at the point n, i.e., xn= f(n).
A sequence { xn } is said to converge to a real number 'l',
" if given any epsi 0 , there exists an n0 in NN such that, for every n= n0 , we have xn in (l - epsi , l + epsi), i.e., given any neighbourhood of l, however small it may be, there exists a stage after which the terms of the sequence lie in that neighbourhood."
If there is no such real number l, then the sequence is going to be divergent.
A series is sum of terms of a sequence. That is if {xn} is a sequence, then the series determined by it is formally written as x1+x2+........ or sum_(n=1)^oo xn . Define Sn = x1 + x2 + .... + xn for every n in NN. Thus we get a sequence { Sn }, called sequence of partial sums of the given series. The given series sum_(n=1)^oo xn is said to converge to a ral number a, if " the sequence of partial sums {Sn} is convergent to l " and we write sum_(n=1)^ooxn = a.
If there is no such real number a, then we say the series is divergent.
So, to talk about convergence of a series, we should know about convergence of the corresponding sequence of partial sums.
Though convergence or divergence of a sequence can be known somewhat easily, the convergence or divergence of a given series is not going to be that easy. There are many tests, which help us to decide whether a series is convergent or not. Among them Ratio test is the foremost thing. It is one of the most easy and useful test about convergence of a series.
Ratio Test for Convergence of a Series :
Intuitively, the infinite sum x1 + x2 +..... is going to be finite if the sequence x1,x2,... is decreasing, that is, for each n, xn xn+1 ,which implies (xn/xn+1) 1. So, intuitively, if the quantity |xn/xn+1| is greater than 1 then the series is going to converge. Ratio test for convergence says the same thing in a mathematical way.
Statement : Let sum_(n=1)^oo xn be a series of real numbers. Let a = lim_(n-oo)| xn/xn+1|. Then,
If a 1, then the series sum_(n=1)^ooxn is convergent.
If a 1, then the series sum_(n=1)^ooxn is divergent.
When a = 1, then the test is inconclusive about the convergence.
Proof :
Assume a 1. Then there exists a rela number s,such that a s 1. Since lim_(n-oo)|xn/xn+1 | = a, there exists an n0, such that for every n= n0 , | xn/xn+1 | s . That is, for all n=n0 , |xn| s |xn+1| .
So by a small work, we get |xn0| sr|xn+r| , that is |xn+r| (1/s)r |xn0|
Since s 1 , 1/s 1. So the geometric series sum_(n=1)^oo (1/s)n is convergent.
Now sum_(n=1)^oo|xn | = | x1 | + | x2 | + ......+ | xn0-1| + sum_(r=1)^oo | xn0+r | Sn0-1 + | xn0 | sum_(k=1)^oo (1/s)r
Where Sn0-1 = sum_(k=1)^(n0-1) | xk |
Since sum_(r=1)^oo (1/s)r is convergent, let it conevrge to b. That is sum_(k=1)^oo(1/s)r = b.
So sum_(n=1)^oo| xn | Sn0 + | xn0 |.b . Hence the given series is convergent.
The case a1 is similar to the above one, as is left as exercise.
Inconclusive nature when a = 1.
1. Consider sum_(n=1)^oo 1. This series is divergent. But in this case a = 1.
2. Consider sum_(n=1)^oo( 1/n2 ) . This series is convergent and in this case also a=1.
By above two examples we can say that when a = 1, then we cannot conclude anything about the convergence of the series.
An Example on Ratio Test for Convergence :
Test convergence of sum_(n=1)^oo ( n! / 5n ) .
Here we have xn = (n!/5n ) . Therefore check that lim_(n-oo) | xn/xn+1| = oo 1.
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Hence we can surely say that the given series is divergent.
Series play an important role in mathematics. Convergence of series play an equally important role. Convergence of series has indeed revolutionised many developments in mathematics. Convergence of series actually convergence of sequences only! ( the sequence of partial sums).
Alternating seriesis a special type series in which the terms are alternating positive and negative. That is, a series sum_(n=1)^ooan is called an alternating series if ai = 0 for every odd i and aj = 0 for every even j . We can put it in a different way. Let un = an if n is even , and
= -an if n is odd .
Then un = 0 , for every n in NN .
So, an = (-1)n-1 un . Thus we get sum_(n=1)^ooan = sum_(n=1)^oo(-1)n-1un .
Thus, alternatively , we can define " alternating series" , as a series of the type sum_(n=1)^oo(-1)n-1 an , where an = 0 .
In this article we will be learning about convergence of alternating series.
Convergence of Alternating Series : Leibniz's Test
Statement : Let { an } be a sequence of non-negative real numbers such that a1 = a2 = . . . . = an = an+1 = . . . . That is the sequence is decreasing. Then the alternating series sum_(n=1)^oo(-1)n-1 an is convergent.
Proof : By convergence of a series, we mean that the sequence sn = a1 - a2 + a3 - . . . . + (-1)n-1 an of partial sums is convergent. So will prove that the sequence { sn } is convergent.
Note that s2n+1 = a1 - a2 + a3 - . . . . + a2n+1 = a1 +(- a2 + a3) + (- a4 + a5) + . . . . . a2n-1) + (- a2n + a2n+1) = a1 .
For every n in NN. [ Since each term in the parentheses is non-positive ]
Also, s2n+1 = (a1 - a2) + (a3 - a4) + . .. .+ (a2n-1 - a2n) + a2n+1 = (a1 - a2) + (a3 - a4) + . . . . +(a2n-1 - a2n) + (a2n+1 - a2n+2) + a2n+3
= s2n+3
So s2n+1 = s2n+3 for every n in NN .
From above two points, the sub sequence {s2n+1 } is increasing and bounded above. So it is convergent, say lim_(n-oo) s2n+1 = s.
We will prove that lim_(n-oo)sn = s. Its enough to show that lim_(n-oo)s2n = s.
Given that an' s are non negative and decreasing, so lim_(n-oo)an = 0. So lim_(n-oo)a2n = 0.
Now, s2n = s2n-1 - a2n . So lim_(n-oo)s2n = lim_(n-oo)( s2n-1 - a2n )= lim_(n-oo)s2n-1 - lim_(n-oo)a2n = s - 0 = s .
Hence lim_(n-oo)sn = s. This implies that given alternating series is convergent.
An Example Showing Convergence of Alternating Series
Test convergence of sum_(n=1)^oo(-1)n-1 (1/n) .
Solution : Firstly, note that 1/1 1/2 1/3 . . . and all the terms are non-negative.So by Leibniz's test , the given alternating series is convergent.
The above example gives an example of a conditionally convergent series. We already know that sum_(n=1)^oo1/n is not convergent and hence sum_(n=1)^oo(-1)n 1/n is not absolutely convergent, but above example shows that it is convergent. So this series forms an example of conditionally convergent series.
by: mathqa
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