math 166 selected solutions to worksheet problems lm...

Math 166 Selected Solutions to Worksheet Problems LM Spring 2019

2/26 Pr 1f ∫dx

x4 + 25x2

This integral is done with partial fraction decomposition. First we factor the bottom

as much as we can. x4 + 25x2 = x2 (x2 + 25). Now we use PFD:

1

x2 (x2 + 25)=A

x+B

x2+Cx+D

x2 + 25

1 = Ax(x2 + 25

)+B

(x2 + 25

)+ (Cx+D)x2

1 = Ax3 + 25Ax+Bx2 + 25B + Cx3 +Dx2

1 = (A+ C)x3 + (B +D)x2 + 25Ax+ 25B.

By comparing coefficients in this equality we get the system of equations

0 = A+ C

0 = B +D

0 = 25A

1 = 25B.

This gives us A = 0, B = 1/25, C = 0, and D = −1/25. Thus out integral becomes∫dx

x4 + 25x2=

∫1/25

x2+−1/25

x2 + 25dx

=1

25

∫1

x2dx− 1

25

∫1

x2 + 25dx

=1

25

(−1

x

)− 1

25

(1

5arctan

(x5

))+ C

= − 1

25x−(

1

125arctan

(x5

))+R.

1

2

3/5 Pr 1a ∫ ∞−∞

dx

(x2 + 9)5/2

The graph of this function is symmetric about the y-axis because of the x2 term i.e. it

is even. Moreover as x→∞ or −∞, 1

(x2+5)5/2→ 1

(x2)5/2= 1

x5, so we suspect that this

integral converges. (Mathematicians refer to this type of function as a bump function

because of the shape of the graph). Let’s evaluate this by first evaluating the improper

integral. We use trig sub and use the substitution x = 3 tan θ, dx = 3 sec2 θdθ

∫dx

(x2 + 9)5/2=

∫3 sec2 θ

(9 tan2 θ + 9)5/2

dθ

=

∫3 sec2 θ

(9 sec2 θ)5/2dθ

=

∫3 sec2 θ

35 sec5 θdθ

=1

34

∫1

sec3 θdθ

=1

34

∫cos3 θ dθ

This is a trig integral we have seen before, so we proceed using the u-sub u = sin θ,

du = cos θdθ and the identity cos2 θ = 1− sin2 θ.

1

34

∫cos3 θ dθ =

1

34

∫ (1− sin2 θ

)cos θ dθ

=1

34

∫ (1− u2

)du

=1

34

(u− u3

3

)+ C

=1

34

(sin θ − sin3 θ

3

)+ C

Since our original substitution was x = 3 tan θ we see that sin θ = x√x2+9

, so we get

∫dx

(x2 + 9)5/2=

1

34

(x√x2 + 9

− x3

3 (x2 + 9)3/2

)+ C

3

Using the fact that our function is symmetric about the y-axis we now rewrite our

integral and plug in the anti-derivative we calculated to obtain∫ ∞−∞

dx

(x2 + 9)5/2= 2 lim

s→∞

∫ s

0

dx

(x2 + 9)5/2

= 2 lims→∞

1

34

(x√x2 + 9

− x3

3 (x2 + 9)3/2

)∣∣∣∣s0

= 2 lims→∞

1

34

(s√s2 + 9

− s3

3 (s2 + 9)3/2

)

= 2 lims→∞

1

34

(1√

1 + 9/s2− 1

3 (1 + 9/s2)3/2

)

= 21

34

(1− 1

3

)=

4

243.

*dusts off hands*

4

3/5 Pr 1b ∫ ∞7

dx√x2 − 6x

For this problem we see that as x→∞, 1√x2−6x →

1√x2

= 1x. Since our functions tends

to a function whose integral is divergent, the integral we care about will probably be

divergent as well. This also tells us that we should compare our integral to something

similar to 1x. We proceed as normal

x2 ≥ x2 − 6x

1

x2 − 6x≥ 1

x2

1√x2 − 6x

≥ 1√x2

=1

x

Since 1x≤ 1√

x2−6x and∫∞7

1xdx diverges by p-test we see that

∫∞7

dx√x2−6x diverges by

comparison test.

It is worth out time to look at ∫ ∞7

dx√x2 + 6x

.

This integral looks almost exactly like the previous integral except we have a +

instead of a −. We suspect that this shouldn’t change much and that this integral

will diverge just like before. However if we use the same comparison test we

encounter a problem!

x2 ≤ x2 + 6x

1√x2 + 6x

≤ 1

x

The inequality is in the wrong direction! This means our comparison test is hogwash.

Instead can we use the following comparison provided x is large enough:

x2 + x2 ≥ x2 + 6x

1√x2 + 6x

≥ 1√2x2

=1√2x.

Now we have 1√2x≤ 1√

x2+6xand comparison test will provide us with divergence once

again.

5

3/5 Pr 1d ∫ 1

0

lnx

x2dx

It is somewhat hard to tell if this integral will converge or not since lnx and x2 behave

very differently as x → 0, but we can try to evaluate this integral. First we use a

t-sub to rearrange terms, let t = lnx, so dt = 1xdx and x = et. This transforms our

integral into ∫ 0

−∞te−t dt = lim

r→∞

∫ 0

−rte−t dt

Now we use integration by parts. Let u = t and dv = e−tdt, so that du = dt and

v = −e−t. With the chosen parts we see that

limr→∞

∫ 0

−rte−t dt = lim

r→∞−te−t

∣∣∣∣0−r

+

∫ 0

−re−t dt

= limr→∞−te−t

∣∣∣∣0−r

+ e−t∣∣∣∣0−r

= limr→∞

(0e0 − rer

)+(e0 − er

)= lim

r→∞−rer + 1− er

= limr→∞−er (1 + r) + 1

= divergent to −∞

The limit diverges because as r →∞, er →∞ and 1 + r →∞.

6

3/5 Pr 1e

∫ ∞1

1− 2 lnx

x3dx

This integral should converge, but it may be hard to see why at first. If we look

at the derivative of the top and bottom functions we see that ddx

(1 − 2 lnx) = −2x

and ddxx3 = 2x2 respectively. This tells us that x3 should increase much faster than

1 − 2 lnx (in fact lnx is typically the slowest growing function one encounters) so

as x → ∞, 1−2 lnxx3

→ 1x3

. Anyway, we proceed with the evaluation of this integral

by u-sub. Let u = 1 − 2 lnx, so du = −2xdx and x2 = e1−u. Plugging this into our

integral we get

∫ ∞1

1− 2 lnx

x3dx = lim

t→∞

∫ t

1

1− 2 lnx

x3dx

= −1

2limt→∞

∫ 1−2 ln t

1

u

e1−udu

= −1

2limt→∞

∫ 1−2 ln t

1

ue−(1−u) du

= −1

2limt→∞

∫ 1−2 ln t

1

ueu−1 du

Now we use IBP similar to the previous problem and this gives us

−1

2limt→∞

∫ 1−2 ln t

1

ueu−1 du = −1

2limt→∞

ueu−1∣∣∣∣1−2 ln t1

−∫ 1−2 ln t

1

eu−1 du

= −1

2limt→∞

ueu−1∣∣∣∣1−2 ln t1

− eu−1∣∣∣∣1−2 ln t1

= −1

2limt→∞

((1− 2 ln t) e1−2 ln t−1 − 1

)−(1− (1− 2 ln t) e1−2 ln t−1

)= −1

2limt→∞

((1− 2 ln t) eln t

−2 − 1)−(

1− (1− 2 ln t) eln t−2)

= −1

2limt→∞

((1− 2 ln t) t−2 − 1

)−(1− (1− 2 ln t) t−2

)= −1

2limt→∞

2 (1− 2 ln t) t−2 − 2

= limt→∞

1− (1− 2 ln t)

t2

7

To evaluate this limit we use L’Hopitals rule because as t→∞ we see that 1−2 ln tt2→

−∞∞ . This gives us

limt→∞

(1− 2 ln t)

t2= lim

t→∞

−2t

2t

= limt→∞

−1

t2

= 0.

Therefore our answer is ∫ ∞1

1− 2 lnx

x3dx = 1.

8

3/5 Pr 3b ∫ ∞1

e−(x+x−1) dx

e−x shrinks incredibly fast because ddxe−x = −e−x, so when we see something like∫∞

ae−x dx we should almost always expect convergence. Here we see that as x→∞,

e−(x+x−1) → e−x so let’s show convergence. We start with x and see that

x ≤ x+ x−1

ex ≤ ex+x−1

1

ex+x−1 ≤1

ex

e−(x+x−1) ≤ e−x.

Now we show that∫∞1e−x dx converges.∫ ∞

1

e−x dx = limt→∞

∫ t

1

e−x dx

= limt→∞−e−x

∣∣∣∣t1

= limt→∞−e−t + e

= e

Therefore since e−(x+x−1) ≤ e−x and∫∞1e−x dx converges,

∫∞1e−(x+x−1) dx converges

by comparison test.

10

3/19 Pr 1a Calculate the arc length of the curve f(x) = 13x3/2 − x1/2 on the interval [2, 8].

This problem involves a nifty trick which is often useful for evaluating integrals of

this nature. We begin by setting up the arc length integral. We have

f ′(x) =1

2x1/2 − 1

2x−1/2

(f ′)2 =

(1

2x1/2 − 1

2x−1/2

)2

=1

4x+

1

4x−1 − 1

2

1 + (f ′)2 =1

4x+

1

4x−1 +

1

2

From here the arc length integral becomes∫ 8

2

√1

4x+

1

4x−1 +

1

2dx

which initially looks bad, but can be dealt with quite easily. Keep in mind that our

ultimate goal is to evaluate this integral, so we need to simplify the term in the square

root to do this. Notice that we can rewrite 1 + (f ′)2 in the following way

1 + (f ′)2 =

(1

2x1/2

)2

+

(1

2x−1/2

)2

+ 2

(1

2x1/2

)(1

2x−1/2

)=

(1

2x1/2 +

1

2x−1/2

)2

.

It will probably be difficult to see this at first, but it is important to keep in mind

that(12x1/2

)2+(12x−1/2

)2+ 2

(12x1/2

) (12x−1/2

)has the form a2 + b2 + 2ab which can

be factored as (a + b)2. If we input this simplification into our integral we now see

that ∫ 8

2

√1

4x+

1

4x−1 +

1

2dx =

∫ 8

2

√(1

2x1/2 +

1

2x−1/2

)2

dx

=

∫ 8

2

1

2x1/2 +

1

2x−1/2 dx

=1

3x3/2 + x1/2

∣∣∣∣82

=

(1

383/2 + 81/2

)−(

1

323/2 + 21/2

)

11

3/19 Pr 2 Compute the surface area of the object obtained by rotating the curve y = 4x+ 3 on

the interval [0, 1] about the x-axis.

This is a typical surface area problem. First we will set up the integral. We have

y′ = 4

(y′)2 = 16

1 + (y′)2 = 17

Thus our integral becomes

2π

∫ 1

0

y√

1 + (y′)2 dx = 2π

∫ 1

0

(4x+ 3)√

17 dx

Evaluation here is fairly easy. We get

2π

∫ 1

0

(4x+ 3)√

17 dx = 2π√

17(2x2 + 3x)

∣∣∣∣10

= 2π√

17(2 + 3)

12

3/19 Pr 3 Use a comparison to show that the arc length of y = x4/3 over [1, 2] is no less than 53.

It is important to keep in mind what we want to get out of this problem, we want to

bound the arc length integral of a particular curve by something. The first step is to

write down the arc length integral we care about. We have

y′ =4

3x1/3

(y′)2 =

(4

3x1/3

)2

=16

9x2/3

1 + (y′)2 = 1 +16

9x2/3

which gives us the integral

∫ 2

1

√1 +

16

9x2/3 dx.

Our goal is to show that

5

3≤∫ 2

1

√1 +

16

9x2/3 dx

by finding a function g with the property that

g(x) ≤√

1 +16

9x2/3.

The idea is that if we integrate both sides of this equation we get

∫ 2

1

g(x) dx ≤∫ 2

1

√1 +

16

9x2/3 dx

and hopefully we found a g so that the first integral is 53. Notice that the function√

1 + 169x2/3 is increasing on [1, 2], in other words we see that

√1 +

16

912/3 ≤

√1 +

16

9x2/3√

25

9≤√

1 +16

9x2/3

5

3≤√

1 +16

9x2/3.

13

Because of this we can let g(x) = 53

so then∫ 2

1

5

3dx ≤

∫ 2

1

√1 +

16

9x2/3 dx

5

3x

∣∣∣∣21

≤∫ 2

1

√1 +

16

9x2/3 dx

5

3≤∫ 2

1

√1 +

16

9x2/3 dx

14

Quiz 7 Find the surface area of the solid obtained when the curve y = 14x2 − 1

2lnx on the

interval [1, 2] is rotated about the x-axis.

A lot of our time will go into the set up of this integral. Let’s find 1 + (y′)2. First we

have

y′ =1

2x− 1

2x−1

(y′)2 =

(1

2x− 1

2x−1)2

=1

4x2 +

1

4x−2 − 1

2

1 + (y′)2 =1

4x2 +

1

4x−2 +

1

2=

(1

2x+

1

2x−1)2

Therefore our surface area integral becomes

2π

∫ b

a

y√

1 + (y′)2 dx = 2π

∫ 2

1

(1

4x2 − 1

2lnx

)√(1

2x+

1

2x−1)2

dx

= 2π

∫ 2

1

(1

4x2 − 1

2lnx

)(1

2x+

1

2x−1)dx

= 2π

∫ 2

1

(1

8x3 +

1

8x− 1

4x lnx− 1

4x−1 lnx

)dx.

We only need power rule to evaluate the first two terms in this integral, so we will

ignore those for now. The last two terms require a bit more work. The term x lnx

can be evaluated with the by parts of u = lnx and dv = x dx. This gives∫ 2

1

−1

4x lnx dx = −1

4

∫ 2

1

x lnx dx

= −1

4

(1

2x2 lnx

∣∣∣∣21

−∫ 2

1

1

2x dx

)

= −1

4

(1

2x2 lnx− 1

4x2)∣∣∣∣2

1

The last term in the integration can be done via a u-sub u = lnx and du = x−1 dx

which gives ∫ 2

1

−1

4x−1 lnx dx = −1

4

∫ 2

1

x−1 lnx dx

= −1

4

∫ 2

1

u du

=1

4

(1

2(lnx)2

)∣∣∣∣21

15

Putting this all together the end result is

2π

∫ 2

1

(1

8x3 +

1

8x− 1

4x lnx− 1

4x−1 lnx

)dx

2π

(1

32x4 +

1

16x2 − 1

4

(1

2x2 lnx− 1

4x2)− 1

4

(1

2(lnx)2

))∣∣∣∣21

.

Which evaluates to some number.

16

3/26 Pr 1e Determine whether the sequence

yn =e−n + (−3)n

5n

converges and if it does find its limit.

We start this problem by splitting it into two parts. We have

e−n + (−3)n

5n=e−n

5n+

(−3)n

5n=

1

en5n+

(−3

5

)n.

Since limits distribute over sums we can compute the limit of each term individually.

We have

limn→∞

1

en5n= 0

for the first term. The second term is alternating so we have to use the squeeze

theorem. We have that

−(

3

5

)n≤(−3

5

)n≤(

3

5

)n.

We also have that 35< 1 so

(35

)n+1<(35

)nand therefore

(35

)nis decreasing. Since

(35

)nis decreasing and bounded below by 0 we have that it converges and one can check

that it converges to 0. Putting it all together we get

−(

3

5

)n→ 0 and

(3

5

)n→ 0

so by squeeze theorem

limn→∞

(−3)n

5n= 0.

Finally we return to our original limit and we see that

limn→∞

e−n + (−3)n

5n= lim

n→∞

e−n

5n+

(−3)n

5n= 0

17

3/26 Pr 1f Determine if the sequence

zn =√n ln

(1 +

1

n

)converges. If it does, find it’s limit.

If we naively take the limit we find that

√n ln

(1 +

1

n

)→ 0 · ∞.

This is an indeterminate form and so we cannot conclude anything yet. Whenever we

see an indeterminate form we should think to try using L’Hopital’s rule. We proceed

by rewriting:√n ln

(1 +

1

n

)=

ln(1 + 1

n

)1√n

→ 0

0.

Now that we have the form 00, we can apply L’Hopital’s rule. Therefore we see that

limn→∞

ln(1 + 1

n

)1√n

= L’Hop = limn→∞

−n−2

1+ 1n

−12n−3/2

= limn→∞

−1n2+n

−12n−3/2

= limn→∞

2

(n2 + n)n−3/2

= limn→∞

2

n1/2 + n−1/2

= limn→∞

2n+1n1/2

= limn→∞

2n1/2

n+ 1

= L’Hop = limn→∞

n−1/2

1= 0

Therefore we can say that

limn→∞

√n ln

(1 +

1

n

)= 0.

(If you think the limit was obvious you should try to do the same thing for n ln(1 + 1

n

).)

18

3/26 Pr 2a Determine whether the following statement is true or false.

If {an} and {bn} are two sequences and the sequence {anbn} converges, then either

{an} or {bn} must converge.

We will show that this statement is false by giving a counterexample. Consider the

sequences an = (−1)n and bn = sin(π(2n+1)

2

). Both an and bn are divergent. We will

write out a few terms to convince ourselves of this.

a0 = 1, a1 = −1, a2 = 1, a3 = −1, · · ·

b0 = 1, b1 = −1, b2 = 1, b3 = −1, · · ·

Notice that if we multiply these sequences together we get

a0b0 = 1, a1b1 = 1, a2b2 = 1, · · · .

In other words our sequence {anbn} is just the constant sequence {1}. Therefore we

have two sequences that do not converge and a product sequence which does converge

which shows that the statement is false.

19

3/26 Pr 5a Show that the sequence

an+1 =√

2 + an with a0 = 0

converges by showing it is bounded above by 2 and increasing.

Recursive sequences require some care to work with, so let’s start by showing it is

bounded above. This means we want to find some number C so that an+1 < C for

every n. Since we are given the bound of 2 we can use that here. If

an < 2

then

2 + an < 2 + 2 = 4.

It follows that

an+1 =√

2 + an < 2.

This tells us that if at least one of the terms in our sequence is bounded above by 2

then all the terms have to be. Since a0 < 2 we have that our sequence is bounded

above by 2.

Now we want to show that this sequence is increasing. ”Increasing” means

an+1 > an

i.e. the n+ 1’th term is always larger than the n’th term. the square root makes this

not so obvious, but with a bit of trickery we can show this. We start with

(an+1)2 − (an)2.

Substituting an+1 we get

(√

2 + an)2 − (an)2 = 2 + an − (an)2 = (2− an)(1 + an) > 0.

From this we can see that

(an+1)2 − (an)2 > 0

so

(an+1)2 > (an)2 ⇒ an+1 > an

From all this work we see that the sequence an converges. (One can also show that

this sequence converges to 2.)

20

4/08 Pr 2c Determine if the series∞∑n=2

1

n(lnn)2

converges or diverges by integral test.

First we need to make sure that we can use the integral test, so we check that 1n(lnn)2

is positive and decreasing. Clearly, 1n(lnn)2

> 0, so it’s positive. To check decreasing

we compute the derivative ddx

1x(lnx)2

= − lnx+2x2(lnx)3

, so it’s decreasing. Now we use the

integral test. ∫ ∞2

1

x(lnx)2dx =

∣∣∣∣u = lnx, du =1

xdx

∣∣∣∣=

∫ ∞ln 2

1

u2du

= −1

u

∣∣∣∣∞ln 2

= limt→∞−1

t+

1

ln 2

=1

ln 2

Therefore since∫∞2

1x(lnx)2

dx converges the series∑∞

n=21

n(lnn)2converges by integral

test.

21

4/08 Pr 2c Determine if the series∞∑n=1

n3

n5 + 4n+ 1

converges or diverges by comparison test.

First we think about what happens to n3

n5+4n+1as n→∞. If n is very large then

n3

n5 + 4n+ 1→ n3

n5=

1

n2

Thus since∑∞

n=11n2 converges our sum should converge as well. Also, from this

reasoning, we want to compare n3

n5+4n+1to n3

n5 .

Now we can start the problem. We have

n5 ≤ n5 + 4n+ 1

1

n5 + 4n+ 1≤ 1

n5

n3

n5 + 4n+ 1≤ n3

n5=

1

n2

∞∑n=1

n3

n5 + 4n+ 1≤

∞∑n=1

1

n2.

Since∑∞

n=11n2 converges by p-test

∑∞n=1

n3

n5+4n+1converges by comparison.

22

4/08 Pr 2d Determine if the series∞∑n=1

sin2 n

n2

converges or diverges by comparison test.

For this problem we need to use the fact that sinn is bounded. Since −1 ≤ sinn ≤ 1

we have that 0 ≤ sin2 n ≤ 1. Thus we have that

sin2 n ≤ 1

sin2 n

n2≤ 1

n2

∞∑n=1

sin2 n

n2≤

∞∑n=1

1

n2.

Since∑∞

n=11n2 is convergent by p-test

∑∞n=1

sin2 nn2 is convergent by comparison.

23

4/16 Pr 1b Determine whether the alternating series∞∑n=1

(−1)n−1

n1/3converges absolutely, condi-

tionally, or not at all.

First we check absolute convergence. Consider∞∑n=1

∣∣∣∣(−1)n−1

n1/3

∣∣∣∣ =∞∑n=1

1

n1/3.

This is a p-series with p = 1/3 < 1. Thus it diverges by p-test. Now we check

conditional convergence. Since this is an alternating series we will use the alternating

series test. We need to show:

(1) that the sequence1

n1/3is decreasing, and

(2)1

n1/3→ 0.

For the first condition we will just take a derivative:

1/n1/3 = n−1/3,

sod

dnn−1/3 = −1

3n−4/3.

Since the derivative is negative the function is decreasing. The second condition is

obvious, i.e.1

n1/3→ 0.

Since both conditions are satisfied∞∑n=1

(−1)n−1

n1/3

converges conditionally by the alternating series test.

24

4/16 Pr 1c

(1) Determine whether the alternating series∞∑n=1

(−1)nn√n2 + 1

converges absolutely, condi-

tionally, or not at all.

First we check absolute convergence. Consider the series∞∑n=1

∣∣∣∣ (−1)nn√n2 + 1

∣∣∣∣ =∞∑n=1

n√n2 + 1

.

Since the highest powers of n’s on the top and bottom are the same (both 1) we

should use divergence test. We have

limn→∞

n√n2 + 1

= limn→∞

n√n2

1√1 + 1/n2

= limn→∞

1√1 + 1/n2

= 1.

Therefore by the divergence test∞∑n=1

∣∣∣∣ (−1)nn√n2 + 1

∣∣∣∣ diverges. Now we check conditional

convergence. Since

limn→∞

n√n2 + 1

= 1

from before, we know that limn→∞

(−1)nn√n2 + 1

6= 0. Therefore

∞∑n=1

(−1)nn√n2 + 1

diverges by divergence test.

25

4/16 Pr 2c Use the root test on the series

∞∑n=1

(1 +

1

n

)−n2

.

We begin with the root test

limn→∞

n

√∣∣∣∣(1 +1

n

)−n2∣∣∣∣ = limn→∞

(1 +

1

n

)−n2

n

= limn→∞

(1 +

1

n

)−n.

Now we compute this limit. Since there is an n in the exponent we will use natural

logs.

limn→∞

ln

(1 +

1

n

)−n= lim

n→∞−n ln

(1 +

1

n

)→ −∞ · 0

This is an indeterminate so we should rewrite this to use L’Hop. Rewriting this we

get

limn→∞

−n ln

(1 +

1

n

)= lim

n→∞

ln(1 + 1

n

)− 1n

→ 0

0

Now we use L’Hop.

limn→∞

ln(1 + 1

n

)− 1n

= ”L’Hop” = limn→∞

1

(1+ 1n)· (−n−2)

n−2= lim

n→∞−

1

(1+ 1n)

1= −1.

Thus we have that

limn→∞

ln

(1 +

1

n

)−n= −1

so

limn→∞

(1 +

1

n

)−n= e−1.

Now we go back and remember why we did all this. We calculated the root test and

found

limn→∞

n

√∣∣∣∣ (1 +1

n

)−n2∣∣∣∣ = limn→∞

(1 +

1

n

)−n= e−1 < 1

Thus since the limit is less than 1 we know that∞∑n=1

(1 +

1

n

)−n2

converges absolutely.

26

4/18 Pr 1a Determine the convergence of∞∑n=1

(−1)n3n

n!

using the ratio test.

To use the ratio test we must calculate the limit

limn→∞

∣∣∣∣an+1

an

∣∣∣∣where an+1 =

(−1)n+13n+1

(n+ 1)!and an =

(−1)n3n

n!, i.e., we compute

limn→∞

∣∣∣∣ (−1)n+13n+1

(n+1)!

(−1)n3nn!

∣∣∣∣ = limn→∞

∣∣∣∣ (−1)n+13n+1n!

(n+ 1)!(−1)n3n

∣∣∣∣.Let’s use the following facts that

(n+ 1)! = (n+ 1)(n)(n− 1)(n− 2) . . . (2)(1) = (n+ 1)n!(1)

3n+1 = 3n3.(2)

Therefore we get

limn→∞

∣∣∣∣ (−1)n+13n+1n!

(n+ 1)!(−1)n3n

∣∣∣∣ = limn→∞

∣∣∣∣ 3n3n!

(n+ 1)n!3n

∣∣∣∣ = limn→∞

∣∣∣∣ 3

(n+ 1)

∣∣∣∣ = 0.

Therefore we have limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = 0 < 1, so by ratio test the series

∞∑n=1

(−1)n3n

n!


27

4/18 Pr 1b Determine the convergence of∞∑n=1

(n!)2

(2n)!

using the ratio test.

Just like before we need the terms an+1 =((n+ 1)!)2

(2(n+ 1))!and an =

(n!)2

(2n)!. We write out

the typical limit

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

∣∣∣∣ ((n+1)!)2

(2(n+1))!

(n!)2

(2n)!

∣∣∣∣ = limn→∞

∣∣∣∣((n+ 1)!)2(2n)!

(n!)2(2(n+ 1))!

∣∣∣∣Now we simplify using fact (1) from (4/18 Pr 1a). Thus we get

limn→∞

∣∣∣∣((n+ 1)!)2(2n)!

(n!)2(2(n+ 1))!

∣∣∣∣ = limn→∞

∣∣∣∣((n+ 1)n!)2(2n)!

(n!)2(2n+ 2)!

∣∣∣∣= lim

n→∞

∣∣∣∣ (n+ 1)2(n!)2(2n)!

(n!)2(2n+ 2)(2n+ 1)(2n)!

∣∣∣∣= lim

n→∞

∣∣∣∣ (n+ 1)2

(2n+ 2)(2n+ 1)

∣∣∣∣= lim

n→∞

∣∣∣∣ n2 + 2n+ 1

4n2 + 6n+ 2

∣∣∣∣= ”L’Hop” = lim

n→∞

∣∣∣∣2n+ 2

8n+ 6

∣∣∣∣= ”L’Hop” = lim

n→∞

∣∣∣∣28∣∣∣∣

=1

4

Therefore limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = limn→∞

=1

4< 1, so by ratio test the series

∞∑n=1

(n!)2

(2n)!


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