Math 237. Calculus II
HW Solutions for 9.4
Assigned: 3, 4, 5, 9, 10, 13, 18, 21, 24, 29, 31, 34, 42, 46
Selected for Grading: 3, 10, 21, 42(c)
Solutions.
3. The series
looks, to me, like the series
, which I know (by the p-series test)
converges. Using the limit comparison test, as instructed, I get that
Since this is positive and finite, and the second series converges, then (by the limit comparison test) the first
series converges.
4. The series
looks to me like the series
Using the limit comparison test, I get that
, which I know (by the p-series test) converges.
So both series converge.
5. Using the ratio test on the series
, I get the limit
So the series converges.
9. Using the ratio test on the series
, I get
So the series converges.
10. The given series is
. Applying the ratio test yields
So the series converges.
13. To determine the convergence or divergence of
, I will use the limit comparison test with
The limit to investigate is
So, by the limit comparison test and the p-series test, the series
converges.
18. I think that the series
converges. I'll try using the ratio test.
So the series converges.
21. The nth term in the series
is
test, with the second series being the convergent (p-series test)
. I'll use the limit comparison
.
So the given series converges.
24. I will determine the convergence or divergence of the series
using the ratio test.
so, by the ratio test, the series converges.
29. Since
for all n ≥ 1 and the series
comparison test, the series
converges (by the p-series test) then, by the ordinary
converges.
31. I will use the ratio test to determine whether
converges.
So the series converges.
34. Since for all n,
and the series
, which is the geometric series with a = r = 1/5,
converges (to 1/4), then by the ordinary comparison test, the series
converges.
42. We aren't asked to prove the root test, just to use it.
(a) To see that the series
converges, note that
is positive for n ≥ 2 and that
. Since 0 < 1 then, by the root test, the series converges.
(b) Since
for all n ≥ 1 and
then the series
converges, by the root test.
(c) Since
for all n ≥ 1 and
converges, by the root test.
then the series
46. Test for convergence or divergence. (These were fun! If anyone got any of these in a way that was simpler
than the way I did, I will give extra credit.)
Recall some stuff from Calculus I. For small positive x-values, the graph of y = sin x is concave down.
That means that it lies below its tangent line at x = 0, which has equation y = x. Here's a picture to show
that.
y
y=x
y = sin(x)

x

Since

for small positive x-values, then
sides of that last inequality shows that
for large integers n. Squaring both
for large integers n. And the p-series
converges. So, by the ordinary comparison test, the series
converges.
I think I can use the same approach for this one as I used for the one in part (a).
First I'll find the equation of the line that is tangent to y = tan x at x = 0:
y = tan x, tan 0 = 0.
y' = sec2x, sec2(0) = 1.
So the equation of the tangent line is simply y = x.
Next I'll check on concavity.
y'' = 2 sec x ∙ sec x tan x = 2 sec2x tan x, which is positive for small positive values of x.
So the graph of y = tan x is concave up for small positive x-values.
Which means that that graph will be above the tangent line for small positive x-values.
Here's a picture to show that.
y
y = tan(x)
y=x

x

So, for small positive x-values, tan x > x.
This means that for large integers n,

.
Since the harmonic series diverges then, by the ordinary comparison test, so does
.
First, for large values of n, 1/n is close enough to zero that we can be sure that
also know that
over itself:
, and so we
. That allows us to multiply the nth term of the series by this quantity
Again, for large n-values, we can assume that cos(1/n) is at least 0.5, and so 1 + cos(1/n) ≥ 1.5. Thus
From what I did above in part (a), for large values of n we can assume that
By the p-series test, the series
series
. So we get
converges and so, by the comparison test, the
does too.