Question

Tell whether the series converges or diverges. If it converges, give its sum

Answer 1

Diverges. Oh, wait...

Answer 2

\[\sum_{n=0}^{\infty}\sin ^{n}(\frac{ \pi }{ 4 }+npi)\]

Answer 3

LOL

Answer 4

What do you glean from the first few terms?
Might it be helpful to expand the sine of a sum?

Answer 5

Ummm

Answer 6

Well the first term is 1

Answer 7

The second is root 2/2

Answer 8

negative root2/2

Answer 9

Okay, that's enough of that.

Remember your trig formula for the sine for a sum? Time to break it out.

\(\sin(\pi/4 + n\pi) = \)

Answer 10

Do you mean the formula for a sine series?

Answer 11

No. Just trigonometry.

\(\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\cos(\alpha)\sin(\beta)\)

Ringing bells? You're supposed to remember this stuff. :-)

Answer 12

Nope. I do remember the book giving us that formula for a later problem though

Answer 13

Actually I do vaguely recall using that formula

Answer 14

Well, there you go. What do you get with that expansion?

Quiet Voice in Your Head: "Remember your trig..."

Answer 15

sin(pi/4)cos(npi)+sin(npi)cos(pi/4)
So (root2/2)cos(npi)+root2/2(sin(npi))

Answer 16

That's probably not what you're looking for though

Answer 17

No, that's perfect. Now, let's think about it.

\(\sin(n\pi) = 0\) for all integers, n, right? We can just throw out that term.

What say you of the other term.

Answer 18

Yes. And cos(npi) oscillates between -1 and 1 for integers

Answer 19

Super. So the actual nth term is \(\left(\dfrac{(-1)^{n-1}}{\sqrt{2}}\right)^{n}\).

Can we conclude anything about that?

Answer 20

How did you get that so quickly?

Answer 21

Oh wait

Answer 22

You just did the a(r)^(n-1) thing and raised it to the nth power because it is the sine of the nth power

Answer 23

I see

Answer 24

So, now what? Is it an alternating series, or not.

\(\left((-1)^{n-1}\right)^{n}\) What can you say about the resolved exponent on -1?

Answer 25

Wait but it is root 2 over 2

Answer 26

Not just root 2

Answer 27

Right, but let's just worry about one piece at a time.

This part is easy: \(\left(\dfrac{1}{\sqrt{2}}\right)^{n}\)

Answer 28

The series alternates between -root2/2 and root2/2

Answer 29

\(\dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}\)!

Answer 30

Are you SURE it alternates? This is my question.

Answer 31

I'm pretty sure

Answer 32

(root2/2 cos(npi))^n

Answer 33

and cos(npi) alternates

Answer 34

Wat

Answer 35

Sorry about that...

n * (n-1) is always odd. Can you prove it? If n * (n-1) is always odd, then it is NOT alternating.

Answer 36

I still dont understand. Where did you get n(n-1)

Answer 37

It's the exponent on -1.

Answer 38

There's no -1 in (root2/2 cos(npi))^n

Answer 39

There was a minute ago. You lost it along the way.

If n is even, then \(\cos(n\pi) = 1\).
If n is odd, then \(\cos(n\pi) = -1\).

This is exactly equivalent to \((-1)^{n-1}\)

Answer 40

There's no way my teacher would expect us to recognize that

Answer 41

That's why you were given the trig expansion.
You already noticed it on your own when you said, "The series alternates between -root2/2 and root2/2".

Answer 42

Yes. I amend that statement to it alternates between (-root2/2)^n and (root2/2)^n

Answer 43

The fact of the matter is, \(\sin(\pi/4 + n\pi) = \left(\dfrac{(-1)^{n-1}}{\sqrt{2}}\right)^{n}\). This is what you just said, agreed?

Answer 44

... Sorry, with the n-exponent on the sine. Typo.

Answer 45

I guess. I dont know where the 2 went

Answer 46

And I dont know where the (-1)^n-1 came from

Answer 47

What 2? It's arithmetic. \(\dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}\)

Answer 48

LOL

Answer 49

Sorry my brain is fried

Answer 50

How did you determine that it was alternating?

Answer 51

Well cosine alternaes

Answer 52

Amend: How did you CORRECTLY determine that it was alternating?

Answer 53

...and that is all I have written. If n is an integer, then \(\cos(n\pi) = (-1)^{n-1}\). Agreed?

Answer 54

Ok

Answer 55

I get that now

Answer 56

So, now we have to buy the whole structure.

\(\sin(\pi/4 + n\pi) = \dfrac{(-1)^{n-1}}{\sqrt{2}}\) \(\huge{IF}\) n is an integer, agreed?

I have deliberately ignored the big exponent at this point. We'll put it back in a minute.

Answer 57

Ok. Is this if n=0 or if n=1? n=1 right?

Answer 58

Certainly for 0 and 1, but also ANY integer. Make sure you believe it.

Answer 59

I meant for sigma notation

Answer 60

Its from n=1 to infinity right?

Answer 61

That's a fair question, but actually no. Let's fix your fears about that.

Is this \((-1)^{n-1}\) ANY differetn from \((-1)^{n+1}\) ?

Answer 62

* different

Answer 63

no

Answer 64

So, if we use +1 instead of -1, do you feel better about leaving it at n = 0 for starters?

Answer 65

Yes

Answer 66

Because its weird to see the n-1 when you are plugging in 0

Answer 67

Fair enough.

Now, the ENTIRE argument.

\(\sin^{n}(\pi/4 + n\pi) = \left(\dfrac{(-1)^{n+1}}{\sqrt{2}}\right)^{n}\).

Do you believe?

Answer 68

Yes

Answer 69

How about this?

\(\sin^{n}(\pi/4 + n\pi) = \dfrac{\left((-1)^{n+1}\right)^{n}}{\sqrt{2}^{n}}\).

Answer 70

Yes

Answer 71

Okay, so the total exponent in the numerator is n*(n+1).

Is n*(n+1) Odd or Even - Always, Sometimes, or Never? Think about the parity of each factor.

Answer 72

OOOH

Answer 73

If n is even, then n+1 is what?
If n is odd, then n+1 is what?

Answer 74

If n is even, n+1 is odd
If n is odd, n+1 is even

Answer 75

So, they NEVER match up. There is always one odd and one even. Agreed?

Answer 76

Yes

Answer 77

What does that mean for the product?

Answer 78

If n is even, the product will be even
If n is odd, the product will be odd

Answer 79

Wait no

Answer 80

If n is odd, the product will be even

Answer 81

Wait I'm wrong again. I go with my original statement

Answer 82

Carefully...

If n is even, n+1 is odd and n*(n+1) is even*odd and that is always odd.
If n is odd, n+1 is even and n*(n+1) is odd*even and that is always odd.

That product is ALWAYS odd. You need to believe this.

Answer 83

The second part is incorrect

Answer 84

3x2=6

Answer 85

Thats even

Answer 86

So, is it ALWAYS even?

Answer 87

I'm talking about the second part of your statement where you said an odd times an even is odd

Answer 88

How about the first part where I said even * odd is odd. Is that also incorrect?

Answer 89

Yes it is

Answer 90

Wait no

Answer 91

Thats correct

Answer 92

Wait no its not

Answer 93

2x3=6

Answer 94

WHAT AM I THINKING

Answer 95

You have it! Don't give up!

Correct, now!

If n is even, n+1 is odd and n*(n+1) is even*odd and that is always even.
If n is odd, n+1 is even and n*(n+1) is odd*even and that is always even.

That product is ALWAYS even You need to believe this.

Answer 96

Ok

Answer 97

One last thing on this.

\(\sin^{n}(\pi/4 + n\pi) = \dfrac{1}{\sqrt{2}^{n}}\)

Do you believe?

Answer 98

I think so?