# calculus II - series comparison test true or false questions - help pls?

### 2 Answers

- PinkgreenLv 77 months agoFavorite Answer
(1) True

(2) True

(3) True

Additional discussion on the " Monster of Math"

(Weierstrass), after reading the message.

The series

cos(xpi3)/2+cos(xpi3^2)/2^2+

cos(xpi3^3)/2^3+...+T(n)+...

where T(n)=cos(xpi3^n)/2^n

The ratio test:

|cos[xpi3^(n+1)/2^(n+1)]*

[2^n/cos(xpi3^n)]

=

|cos[3xpi(3^n)]/[2cos(xpi3^n)|

=

|[4cos^3(xpi3^n)-3cos(xpi3^n)]/

cos(xpi3^n)|/2

=

|4cos^2(xpi3^n)-3|/2

<=

|4-3|/2=1/2<1 (independent of x).

=> the series converges absolutely.

In fact, for n=1 to infinity,

SIGMA[|cos(xpi3^n)/2^n|]<=

SIGMA[1/2^n]. By the comparison test,

the latter is a G.P. with the ratio=1/2 is

convergent, =>

the series is uniformly convergent with x

over D(-inf., inf.), as n->infinity.

Thus, we can write

..........inf.

f(x)=SIGMA[cos(xpi3^n)/2^n]

.........n=1

But, f(x), strictly speaking is a function of x

but it is not an injection & contains the

"many to one" types of corresponding. For

any x1& x2 in D, if

cos[(x1)pi3^n]=cos[(x2)pi3^n]

=>

cos[(x1)pi3^n]-cos[(x2)pi3^n]=0

=>

-2sin[pi3^n(x1+x2)/2]*

sin[pi3^n(x1-x2)/2]=0

=>

sin[pi3^n(x1+x2)/2]=0 or

sin[pi3^n(x1-x2)/2]=0

=>

x1=2k/3^n-x2 or

x1=2k/3^n+x2

for some k & n, but not exactly x1=x2.

The graph of it for n=20, is very complicated

like the "sound wave" but is continuous over

there. It may expect that f(x) is continuous

over D too.

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