INTEGRAL TEST AND RATIO TESTMATHEMATICAL PHYSICS

Written by :

1. Laila Nur Afianti K23100582. Okki Wahyu Priutami K2310072

Physic Education

FACULTY OF TEACHERSHIP AND EDUCATION SCIENCE

SEBELAS MARET UNIVERSITY

SURAKARTA

INTEGRAL TEST

We can use this test when the term of the series are positive and not increasing, that is

when an+1≤an . (Again remember that we can ignore any finite number of term of the series,

thus the test can be used even if the condition an+1≤an does not holt for a finite number of

terms.) To apply the test we think of an as a function of the variable n, and, forgetting our

previous meaning of n, we allow it to take all values, not just integral ones. The test states

that:

If 0<an+1<an for n > N, then ∑∞

an converges if ∫∞

andn is finite and diverges if

the integral is infinite. (The integral is to be evaluated only at the upper limit; no lower limit

is needed.)

To understand this test, imagine a graph sketched of an as a function of n. for

example, in testing the harmonic series ∑ n=1∞ 1/n , we considered the graph of the function

y=1/n (similar to figures 1 and 2) letting n have all values, not just integral ones. Then the

value of y on the graph on n= 1, 2, 3,…, are the term of the series. In figure 1 and 2, the areas

of the rectangles are just the terms of the series. Notice that in figure 1 the top edge of each

rectangle is above the curve, so that the area of the rectangles, is greater than the

corresponding area under the curve. On the other hand, in figure 2 the rectangles lie below

the curve. Now the areas of the rectangles are just the terms of the series, and the area under

the curve is an integral of y dn or an dn. The upper limits on the integrals is and the lower

limit could be made to correspond to any term of the series we wanted to start with. For

example (see figure 1), ∫3

∞andn

is less than the sum of the series from a3 on, but (see figure

2) greater than the sum of the series from a4 on. If the integral is infinite, then the sum of the

series from a4 on is finite, that is, the series converges. Note again that the terms at the

beginning of a series have nothing to do with convergence. On the other hand, if the integral

is infinite, then the sum of the series from a3 on is infinite and the series diverges. Since the

beginning terms are of no interest, there is no need to use a lower limit on the integral, and

you should simply evaluate ∫∞

andn .

FIGURE 1

FIGURE 2

y= an

y= an

Example : Test for convergence the harmonic series

(1) 1+ 1

2+ 1

3+ 1

4+.. ..

Using the integral test, we evaluate

∫∞ 1

ndn=ln n|∞=∞ .

(We use the symbol ln to mean a natural logarithm, that is, a logarithm to the base e.)

Since the integral is infinite, the series diverges.

RATIO TEST

The integral test depends on your being able to intergrate an dn; this is not always easy! we

consider another test which will handle many cases in which we can not evaluate the integral.

Recall that in thr=e geometric series each term could be obtained by mulltipyling the one

before it by the ratio r, that is, an+1=ran or an+1 /an=r . For the other series the ratio

an+1 /an is not constant but depends on n ; let us call the absolute value of this ration ρn . Let

us also find the limit (if there is one) of ρn as n→∞ and call this limit ρ . Thus we define

ρn and ρ by the equation

ρn=|an+1

an|,

ρ=limn→∞

ρn .

If you recall that a geometric series converges if |r|<1 , it mayy seem plausible that a series

with ρ<1 should conveerge and this is true. This statement can be proved (problem 30) by

comparing the series to be tested with a geomwtric series. Like a geometric series with |r|<1

, a series with ρ<1 also diverges (problem 30). However, if ρ=1 converge and diverge, so

we must find another test (say one oof the two preceding tests). To summarize the ratio test :

ρ < 1, the series converge

If ρ = 1, use a difference test

ρ > 1, the series diverge

Test for conveergence the series

1+ 12+ 1

3+. . .+ 1

n !+. ..

Using (6.2), we have

ρn=|1(n+1) !

÷1n !

|=n!(n+1) !

=n(n−1)(n+1 )(n)(n−1)

=1n+1

ρ=limn→∞

ρn=limn→∞

1n+1

=0

Since ρ<1 , the series converges

Test for convergence the harmonic series

1+ 12+ 1

3+. . .+ 1

n+. ..

We find

ρn=|1(n+1)

÷1n|=n

(n+1)=n

(n+1 ),

ρ=limn→∞

nn+1

=limn →∞

1

1+1n

=1

Here the test tells us nothing and we must use some diifferent test. A word a warning from

this example: Notice that ρ=n(n+1 ) is always less than 1. Be careful not to confuse this

ratio with ρ and conclude incorrectly that this series converges. (it is actually divergent as

we proved by integral test.)remember that ρ is not the same as the ratio ρn=|an+1/an| , but

is the limit of this ratio as n→∞ .

PROBLEM :

For example:

1. Suppose we calculate the value of series, use the integral test to find whether

converge or diverge.

S=1+ 123

+ 133

+ 143

+.. .=∑n=1

∞ 1n3

With calculator we get the partial value until sixth series equal 1,2902916. So, remainder is

S-S6=∑n=7

∞ 1n3

So, it’s a series with integral that limit from x=7 until x=, obtain relation

∫7

∞ 1x3

dx<S−S6<∫7

∞ 1x3

+ 173

Or

0,010204 < S- S6<0,010204 + 0,00291545

Remember that S6=1,2902916 in above, we’ll get

0 , 010204+1 , 2902916<S<0 ,010204+0 , 00291545+1 ,2902916

Or after pembulatan can we write

1,2005< S < 1,2034

So, it’s convergen

2. Use the ratio test to finde whether the following series converge or diverge

a.ρ=∑ ( 4+2 n

7+3 n )n

Answered :

ρ=∑ (4+2 n7+3 n )

n

¿ limn→∞

4+2 n7+3 n

¿

4n

+2 nn

7n

+3nn

¿0+20+3

¿23

Because lim

n→∞

4+2 n7+3n

<1, so this series is converge

b.ρ=∑ ( 4+5n

7+3n )n

Answered :

ρ=∑ (4+5n7+3n )

n

¿ limn→∞

4+5 n7+3 n

¿

4n

+5 nn

7n

+3 nn

¿0+50+3

¿53

Because lim

n→∞

4+5 n7+3 n

>1, so this series is diverge

Daftar Pustaka

Boas, Marya L. 1966. Mathematical Methods in the Physical Sciences. USA: Courier

Compenies, Inc.

Soedojo, Peter. 1995. Asas-Asas Matematika Fisika dan Teknik. Yogyakarta: Gajah Mada

University Press.