2—Infinite Series 27The required observation is that an increasing sequence of real numbers, bounded above, has a
limit.
After some point,k=M, all theukandvkare positive anduk≤vk. The suman=
∑nMvk
then forms an increasing sequence of real numbers, so by assumption this has a limit (the series
converges). The sumbn=
∑nMukis an increasing sequence of real numbers also. Becauseuk≤vk
you immediately havebn≤anfor alln.
bn≤an≤ lim
n→∞an
this simply says that the increasing sequencebnhas an upper bound, so it has a limit and the theorem
is proved.
Ratio Test
To apply this comparison test you need a stable of known convergent series. One that you do have is
the geometric series,
∑
kx
kfor|x|< 1. Let thisxkbe thev
kof the comparison test. Assume at leastafter some pointk=Kthat all theuk> 0.
Also thatuk+1≤xuk.
ThenuK+2≤xuK+1 and uK+1≤xuK gives uK+2≤x^2 uK
You see the immediate extension is
uK+n≤xnuK
As long asx < 1 this is precisely set up for the comparison test using
∑
nuKx
nas the series thatdominates the
∑
nun. This test, theratio testis more commonly stated for positiveukas
If for largek,
uk+1
uk
≤x < 1 then the series
∑
uk converges (2.8)
This is one of the more commonly used convergence tests, not because it’s the best, but because it’s
simple and it works a lot of the time.
Integral Test
The integral test is another way to check for convergence or divergence. Iff is adecreasing posi-
tivefunction and you want to determine the convergence of
∑
∫∞ nf(n), you can look at the integral
dxf(x)and checkitfor convergence. The series and the integral converge or diverge together.
1 2 3 4 5
f(1)
f(2)
f(3)
f(4) f(x)
From the graph you see that the functionf lies between the tops of the upper and the lower
rectangles. The area under the curve offbetweennandn+ 1lies between the areas of the two
rectangles. That’s the reason for the assumption thatfis decreasing and positive.
f(n). 1 >
∫n+1n