Math 166 Selected Solutions to Worksheet Problems LM...
Transcript of 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.
9
3/5 Pr 3c ∫ 1
0
| sinx|√x
dx
For this integral we need to use a fact that may have been forgotten about. This fact is
that | sinx| ≤ x so long as 0 ≤ x. This can be seen by showing that f (x) = x−| sinx|is increasing and f (0) = 0 i.e. the distance between x and | sinx| increases as x
increases. Therefore we have
1√x
=1√x
| sinx|√x≤ x√
x=√x
Now we show that∫ 1
0
√x dx converges∫ 1
0
√x dx =
∫ 1
0
√x dx
=2x3/2
3
∣∣∣∣10
=2
3
Therefore∫ 1
0| sinx|√
xdx converges by comparison test since | sinx|√
x≤√x and
∫ 1
0
√x dx
converges.
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!
converges absolutely.
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)!
converges absolutely.