Question

show that if \(\lim a_{n} = A\) exists, then \(\lim \frac{1}{n} (a_1 + a_2 + ... + a_n) = A\), but that the converse is false

Answer 1

Seems a bit of a strange statement to me, did they specify any limits on the expression? If not I would try something like:
\[\Large \lim a_n=A \Rightarrow \lim a_1=A \Rightarrow \lim a_2=A \dots \]

Answer 2

that would leave an n-times multiple of A, which would be right for the latter statement, but it doesn't seem intuitive to me without restrictions on the limits.

Answer 3

oh ok i get the first part now! but how do i prove the converse is false?

Answer 4

With a counterexample. May be using that the serie doesn't have absolute values.

Answer 5

i'm not sure what that means... sorry

Answer 6

Is possible that we can check
\[a_n=(-1)^n\]
as a counterexample?

Answer 7

But that wouldn't disprove the original statement though?

Answer 8

Converses are always a fun thing to mess with, by logic they are by default false, if they are true then it's an exception from the general case :)

\[\Large p \Rightarrow q \not\Leftrightarrow q \Rightarrow p\]
However you could try be rewriting the expression as follows, assume that q is true, therefore you can write:
\[\Large \lim \frac{1}{n}(a_1+a_2 +a_3+ \cdots +a_n) = \lim \frac{1}{n} \sum_{i=1}^na_i=A \]
If this statement holds true then it most be possible to show that p (our original statement) is true from this as well

Answer 9

but i have to disprove the second indication though. would it be easier for me to just use a counter example like John suggested?

Answer 10

The converse would read as if \( \lim\frac{1}{n}(a_1+a_2+a_3 + \cdots +a_n)=A \) exists, then \( \lim a_n=A\) holds true (as well) . A counter example is really a good idea to continue from here I agree.

As far as it comes to, I would have said that the sum defined in the converse, doesn't give sufficient information about the individual limits of a single term, such as \(a_1\) likewise for \(a_2\) and all the latter terms up to \(a_n\).

Answer 11

ok. for the first part: it says to prove the converse is false, which means I need to prove that if \(\lim \frac{1}{n}(a_1+a_2+...+a_n)=A\) holds, then \(\lim a_n=A\) does not hold?
And then I'm kind of lost when I read the second part...

Answer 12

I wouldn't see that way, but in this other way. For the first part,
\[\exists\lim_{n}a_n=A\Rightarrow\lim_{n}(a_1+a_2+\ldots+a_n)=A\]
For the second,
\[\lim_{n}(a_1+a_2+\ldots+a_n)=A\text{ does not imply }\exists\lim_{n}a_n=A\]

Answer 13

you missed a 1/n in both statements but I get what you're saying.

Answer 14

Yeah, sorry.

Answer 15

so in order to prove the second part, i can just use that counter example? but how would i prove \(\lim \frac{1}{n} ((-1)^1+(-1)^2+...+(-1)^n)\) converge to a number? and what would that number be?

Answer 16

Is this considered to be the limit of a sequence? Because it seems to me like the problem suggest that the limit will always exist, no matter which limit of interest we force upon it, such that it can read as:
\[\lim_{n \to \infty}a_n=\lim_{n \to 0}a_n \] Or does anyone know yet what they want to suggest with the notation by introducing a limit without bounds?

Answer 17

I assume is the limit in infinite.

Answer 18

infinity would be assoumed in this case. sorry.

Answer 19

So for the converse it could read as \[\Large \lim_{n \to \infty} \frac{1}{n}\sum_i^na_n=A=0 \] if I am not mistaken from the notation.

Answer 20

if that's the case, then how would the first part hold?

Answer 21

This is not the case, as a_n can take any value.

Answer 22

but \(\lim \frac{1}{n}=0\)

Answer 23

Example,
\[a_n=n\Rightarrow \lim_{n}(1/n)\sum_{k=1}^na_n=\infty\]

Answer 24

ok but i still don't know where to start so i can prove the converse relationship would be false

Answer 25

Corrected,
\[a_k=k\Rightarrow \lim_n(1/n)\sum_{k=1}^na_k=\infty\]

Answer 26

You need to find a convergent,
\[\lim_{n}\sum_{k=1}^na_k=A\]
But when you evaluate the n-ism term it has no limit.

Answer 27

so would your example of (-1)^n work?

Answer 28

because I dont think it's converging

Answer 29

I have some doubt, but it's true that satisfies that the n-ism term is not convergent. The sum of terms, however, can be rearranged in order to get always zero. Ok, I would take it as counterexample.

Answer 30

i have to go to my psych class thanks guys. can i leave this problem open or should i reopen one after i come back?

Answer 31

@John_ES, could you explain your limit to me of your example? because I reasoned my statement through L'Hopital's Rule which might have been a wrong place to start from at my point:
\[a_k=k\Rightarrow \lim_{n\to \infty}(1/n)\sum_{k=1}^na_k=\infty \]

Assuming that k is any integer, but in general that \(a_k\) is a numeric value, that would mean that for big \(n\) we have something in the form of \(\infty / \infty\), differentiation would lead to 0/1.

I might have mislead myself though with my reasoning.

Answer 32

I think my first part of proof is wrong... i'll be back soon. thanks guys.

Answer 33

Thanks for the problem @johnny0929, it's interesting :)

Answer 34

i'll leavve this open just to see if we can come up with some other explanations lol.

Answer 35

@Spacelimbus, this is the reasoning I take,
\[\sum_{k}^na_k=\sum_{k}^nk=\frac{(1+n)n}{2}\]Then,
\[\lim_n\frac{1}{n}\sum_{k}^na_k=\lim_n\frac{1}{n}\frac{(1+n)n}{2}=\infty\]

Answer 36

absolutely correct @John_ES, thank you, I just got to that point myself by now while rewriting it with the gauss property :-)

Answer 37

I focused too much on the (1/n) and it's behavior for big values of n, but Gauss Addition gets rid of it.

Answer 38

Ok, this counterexample is correct,
\[a_n=(-1)^n\]\[\nexists\lim_na_n\]\[\lim_n[(\sum_k^na_k)/n]=-0.693147\]

Answer 39

In order to get the last limit, you can take the McLaurin series of Ln(1+x),
\[\ln x=x-x^2/2+x^3/3-x^4/4+\ldots=-(-x+x^2/2-x^3/3+x^4/4+\ldots)\]So, for x=1,
\[-\ln2=\sum_n(-1)^n/n\]satisfying that the limit of a_n doesn't exist.

Answer 40

Thanks for all of your help! I got it now thanks!