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From the Ch. 10 Notes:

Integral Test Given1n

nt

=

If

1

ndnt

converges then series converges.

Integral Test: Iffis a continuous, positive, decreasing function on ),1[ with nanf =)( , then the

series

=1n

na converges if and only if the improper integral

1

)( dxxf converges.

So essentially, if the area under a curve from 1 to infinity is finite (the integral converges/exists), thenthe sum of the series also exists.

This should make sense because an integral is the exactarea under the curveit is the summation of

infinitely many tiny rectangles (of width dx). (Think back to the Riemann sums; they were rectanglesunder the curve and then to find the exact area, we allowed the number of rectangles to approach

infinity.)

In a sum however, we only have a rectangle every integer numbereach rectangle is only 1 wide. Thisis because the sigma symbol just goes by integersie 1, 2, 3, etc. It never has 1.1 or 1.5. So because we

have fewer rectangles, it would follow that this sum would be slightly less than the integraltherefore,if the integral exists, than the sum will also exist.

To determine if the integral converges:This goes back to section 8.8improper integrals

As we learned,1

xp1

+

dx converges for values ofp >1 and diverges for values ofp 1.

So if we were given1

xx=1

, we could test this by seeing if the integral converges. In1

xdx

1

,p=1

(because x is to the first power). 11 so the integral diverges and therefore the sum also diverges.

Sometimes you need to combine powers from the numerator and the denominator to find what p is for

the equation. If you are given2x

31

(x21) x

2 4( )(x 2)x=1

, you would rewrite with an integral (you can

essentially skip the step of rewriting for a polynomial expression like this), add the degrees of x in thenumerator and in the denominator independently then add/subtract. In this case, the numerators degree

is 3, and the denominators degree is 5 (because you have an x2

* x2

* x). Subtract 5-3, and you getp=2.

You could also look at this asx

3

x5

which would simplify to1

x2

, also giving youp=2. 2>1, so the series

would converge.

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Some more involved problems:

1

klnkk=2

(k changes to x just because thats convention; you couldleave it as k if you want)

(2 to infinity because our initial term was k=2)

Note that limits of integration change because of the u-sub

Similar problem:

2

n(lnn)3

n=2

2

x(lnx)

3dx

2

u = lnx

du =1

xdx

2

xu3

du1x

ln2

=2

u3ln2

p = 3 and 3 >1; converges

Note that evaluating the integral would NOT give you the answer to the sum; it would be an over

approximation. (We havent learned any way so far to find the sum of this question)

1x lnx2

u = lnx

du = 1xdx

1

xu

du1

xln2

=1

uln2

p = 1 and 11; diverges