Question

HELP PLEASE

Answer 1

I need help with Number 1. :/

Answer 2

@518nad @mathmate @zepdrix

Answer 3

Would it be convergent, since the numbers are gradually getting closer and closer to zero?

Answer 4

Although, there is a up and down flux on the graph.

Answer 5

Are you familiar with geometric series?

Answer 6

Yes.

Answer 7

Can you write the first series in terms of the first term, and the general term an?

Answer 8

uh.....

\[\sum_{10}^{\infty}a_n\]

Answer 9

Well, the first term is definitely 10.
In a geometric series, we get the subsequent terms by multiplying by a constant, r, or the common ratio.
What do we need to multiply 10 to get the next term -8? This is another way to calculate the common ratio, r.

Answer 10

-4/5?

Answer 11

exactly! r=4/5 is the common ratio.
In geometric series, all we need to get started is to find
1. a=first term
2. r=common ratio.
to reproduce the whole series.
This can be written as
Sum of the series = \(\sum_1^{\infty}a_i r^i\)
and a1 is usually written as a.
Ok so far?

Answer 12

Yessir.

Answer 13

Now if the sum is a finite number, the series is convergent. If the sum S is \(+\infty\) or \(-\infty\), the series is divergent.
So far so good?

Answer 14

Yessir

Answer 15

Now how do we find the sum of the series?
There is a magic formula which is worth knowing.
Sum of series, S = a(1-r^n)/(1-r)
where n is the number of terms we are summing.

Answer 16

Let's take an example, trying to sum 1+1/2+1/4+1/8.....
Can you find a and r for this series?

Answer 17

a = 1
r = 1/2

Answer 18

My professor taught me that to find the sum we use... \[Sum = \frac{ a }{ 1-r }\]

Answer 19

Yes, that is correct if we sum to infinite number of terms, because
r^(inf)=0, so that leaves a/(1-r).
Ok, since you are familiar with that, can you find the sum for the example series to infinite number of terms?

Answer 20

You have already found a and r, so you can substitute in your professor's formula.

Answer 21

Sorry im confused...
How does r^(inf)=0?
doesnt that equal infniity?

Answer 22

If r<1, say, 0.99, then r gets progressively smaller and eventually approaches as n-> infinity. So the series converges, i.e. IF r<1.
If r>1, say 1.02, then r gets progressively larger and larger, and eventually approaches infinity. In this case (r>1) the series diverges.
So far so good?

Answer 23

Your professor has probably mentioned that WHEN r<1, then sum = 1/(1-r).

Answer 24

Oh okay, I get it.

Answer 25

So can you find the sum when a=1, and r=1/2, please?

Answer 26

And in this case r is less than 1
since r=1/2

Answer 27

yes, that's correct. (so the series converges). Therefore we can find the sum.

Answer 28

Yes.\[Sum = \frac{ 1 }{ 1-\frac{ 1 }{ 2 } } = 2\]

Answer 29

Exactly.
So let's go back to the first question, which turns out to be an infinite series (i.e. we sum to infinity).
We'll first find a and r then see if you can use your professor's formula.
Can you find a and r for
10, -8, 6.4, -5.12, 4.096.....

Answer 30

So then...\[a=10, r=- \frac{ 4 }{ 5 }\]
\[r=-\frac{ 4 }{ 5 }\rightarrow \left| -\frac{ 4 }{ 5 } \right|\rightarrow \frac{ 4 }{ 5 } >1\]

\[Sum = \frac{ 10 }{ 1+\frac{ 4 }{ 5 } } = \frac{ 10 }{ \frac{ 9 }{ 5 } } = \frac{ 50 }{ 9 }\]?

D:

Answer 31

Perfect! So you just answered the first question. Well done!
Now can you please try the second one? It is awfully similar to the first one.

Answer 32

Oh I got the second one already. 1/17

But for the third one, r = -4/3, after absolute value its 4/3. But thats > 1

Answer 33

What does that tell you then?

Answer 34

That its divergent?

Answer 35

\[\frac{ 4 }{ 5 } <1 \]* typo

Answer 36

Oh oops.

Answer 37

Not really.
There is a little trick you can do.
You can adjust the first term.
Thanks @Nnesha.... She told me that the common ratio should be 8/9 and not 4/3.
But you still can adjust the first term so that the nth term is (8/9)

Answer 38

How did you get that? D:

Answer 39

what?? :o

Answer 40

Actually, you can sum the series as is and get the right answer.
if we write it out, it would be
1/9, 8/81, 64/729, ....
Can you find the first term a and the common ratio r?

Answer 41

r = 8/9

Answer 42

and the first term?

Answer 43

1/9

Answer 44

Good. Will it converge?

Answer 45

yes

Answer 46

And the sum is?

Answer 47

1/17.

But i was talking about number 3

Answer 48

I got that a=8 and r= -4/3
for #3

Answer 49

Ok, I didn't even know there is a #3. We'll get to that,but you probably have the answer.
The sum 1/17 is not correct for #2, you'd have to try again!

Answer 50

Hint: review your professor's formula!

Answer 51

But my homework accepted 1/17 as the answer D:

Answer 52

Sorry, I didn't see the negative sign before the eight. Yes, then r=-8/9, then sum is 1/17.
For #3, what do you say? a, r, convergence, sum?

Answer 53

Well, a = 8 and r = -4/3
4/3 is not less than 1.

So it is divergent?

Answer 54

actually, r=-4/3, and |r|>1 so it is divergent. Correct.

Answer 55

Actually you're more accurate, |r| not less than 1 is the right criterion.

Answer 56

Thanks, I think i get it now. :)

Answer 57

Perfect! You're welcome!
If you have time, there are useful examples here:
http://www.purplemath.com/modules/series5.htm