Question

How do I show a series is convergent and what it converges to?

The series is \[1 + \frac{x^n}{n!} ...\]

Answer 1

You mean
1 + x + x^2/2!d + x^3/3! + x^4/4! + ...
?

Answer 2

Yea

Answer 3

You mean
1 + x + x^2/2! + x^3/3! + x^4/4! + ...
?

Answer 4

There are lots of convergence tests. Which ones do you know so far?

Answer 5

I don't know any. We're supposed to figure it out on our own but I don't really understand

Answer 6

Well, when you have a geometric progression, when does this converge:
a + ar + ar^2 + ar^3 + ar^4 + ar^5 + ... ?

Answer 7

What are 'a' and 'r'? Both variables?

Answer 8

Yes, you recognize this as a geometric series, yes? Such as
1 + 1/2 + 1/4 + 1/8 + 1/16 + ...

in which case a = 1 and r = 1/2

Answer 9

Oh, yea. Convergence just means that the entire series can be represented by a number, right?

Answer 10

Yes, it means
1 + 1/2 + 1/4 + 1/8 + 1/16 + ...
gets arbitrarily close to a given number if we have enough terms of the sum. In this case, the sum converges to 2.

Answer 11

I have that memorized but I forgot why it's true or the proof for it

Answer 12

Given that you are having trouble with a geometric series, I have a hard time that you are being asked to come up with your own convergence test. Are you sure you haven't missed a class/lecture which you need to look up in your text book?

Answer 13

...have a hard time believing that ...

Answer 14

Our last lesson was on inverse functions. Our teacher makes us learn the lesson before he actually teaches it. I don't think we cover series until the very end of the semester because we have to go over more functions and stuff

Answer 15

Well, the proof is instructive:
The finite sum
a + ar + ar^2 + .... + ar^n
is equal to
a (1 - r^(n+1)) / (1 - r)

Look familiar?

Answer 16

I've seen it before

Answer 17

Now what is the limit of r^(n+1) as n --> infinity ?

Answer 18

Doesn't it depend on whether r is <1, =1, or >1

Answer 19

Yes: if |r| < 1, |r| = 1 or |r| > 1.

For what values of r does the limit r^(n+1) exist as n --> infinity?

Answer 20

I think all real numbers

Answer 21

Oh wait

Answer 22

No. If r = 1, for example with a = 1, we have
1 + 1 + 1 + 1 ....
which does not converge.

In other words
\[ S_n = a \frac{1-r^{n+1}}{1-r} \]
Does not converge when n --> infinity

Answer 23

i don't think it has a limit if the number fluctuates

Answer 24

Oh

Answer 25

Why doesn't limit 1^(n+1) as n approaches infinity = 1?

Answer 26

In fact, S_n as we're written it doesn't even exist when r = 1 because the denominator is zero.

But yes indeed,
\[ \lim_{n \rightarrow \infty} r^{n+1} \]
exists provided \( -1 < r \leq 1 \)

But now we just saw that if r = 1, this is a special case and it is clear that if a is not zero, then
a + ar + ar^2 + ar^3 + ....
doesn't exist.

Hence in summary...

Answer 27

The infinite sum
a + ar + ar^2 + ar^3 + ...
exists if and only if |r| < 1

Yes?

Answer 28

Yes

Answer 29

Now, label each of these terms like this
\[ a_0 = a \]
\[ a_1 = ar \]
\[ a_2 = ar^2 \]
...
\[ a_j = ar^j \]
...

Then it is clear that for all \( j > 0 \), \( a_j/a_{j-1} = r \)

Answer 30

Therefore by comparing a sum with a geometric sum,
\[ a_0 + a_1 + a_2 + .... + a_j + .... \]
converges if and only if
\[ \lim_{n \rightarrow \infty} \frac{a_n}{a_{n-1}} < 1 \]

Answer 31

In the case of the series you wrote down
\[ a_0 = 1, a_1 = x, a_2 = x^2/2!, ..., a_j = x^j/j!, ... \]

Answer 32

Therefore
\[ 1 + x + x^2/2! + .... = a_0 + a_1 + a_2 + ... \]
converges if and only if
\[ \lim_{n \rightarrow \infty} \frac{a_n}{a_{n-1}} < 1 \]
i.e.,
\[ \lim_{n \rightarrow \infty} \frac{x^n/n!}{x^{n-1}/(n-1)!} \ = \ \lim_{n \rightarrow \infty} \frac{x}{n} \ < \ 1 \]

Answer 33

As this is true for all real numbers x, it must be that
1 + x + x^2/2! + x^3/3! + ...
converges for all real numbers x

Answer 34

Wow, that is pretty cool

Answer 35

...and not at all obvious if you've not seen it before. I'm surprised you're asked to figure this out.

Answer 36

Now the second part: what is this equal to?

Call this function f(x)
f(x) = 1 + x + x^2/2! + x^3/3! + ...

Now what is df/dx ?

Answer 37

df/dx is 0 + 1 + x + x^2/2 + x^3/6, etc. or the same thing as f(x) I guess? since 6 = 3! I kind of see where this is going

Answer 38

Yes,
f' = f

What function behaves like this?

Answer 39

e

Answer 40

e^x

Answer 41

Oh right

Answer 42

But why did you differentiate to find out what the value it converged to was?

Answer 43

There's one little twist. So also does
g(x) = Ae^x
for any constant A.

How do we get A = 1?

Answer 44

Why? Because it's the easiest method right now. You'll see later how to start with a function and arrive at a function.