Value of alternating sum at infinity

1. May 8, 2017

vishnu 73

i showed that an 2 = a02 if n is even and 1 - ao2 if n is even
how am i supposed to evaluate this at infinity

2. May 8, 2017

Staff: Mentor

I moved your thread to our homework section.
That is not correct (I guess the last word is supposed to be "odd", but both formulas don't work for either case).

3. May 8, 2017

stevendaryl

Staff Emeritus
How did you show that?

Anyway, suppose that $a_0, a_1, a_2, ...$ converges to some number $\alpha$. Then that means that
$a_0 - \alpha, a_1 - \alpha, a_2 - \alpha...$ converges to zero.

So, let's define a new sequence $b_n = a_n - \alpha$. Using the recurrence relation

$a_{n+1} = \sqrt{1-a_n}$

see if you can get an equation for $\alpha$ that involves $b_{n}$ and $b_{n+1}$. Then take the limit as $b_n \rightarrow 0$ and $b_{n+1} \rightarrow 0$

4. May 8, 2017

BvU

IF the limit exists, what do you know about ${a_{n+1}\over a_n}$ for very big values of $n$ ?

5. May 8, 2017

vishnu 73

@mfb
i dont understand why you say that this is what i did
an+1 2 + an2 = 1
hence
first is
a02 next a12 is 1 - a02 then a22 is a02 isn't this alternating please point out my mistake to me
@stevendaryl
thats how i showed to both of you i am very sorry if am doing something wrong
when you say define new sequence as bn = an -a what is the a is it the a0 or an-1 thanks
@BvU
the limit approaches one but has my method been correct so far as others have pointed something is amiss
if you too could clarify that then i would gladly work along the line

thanks to all!!!

6. May 8, 2017

stevendaryl

Staff Emeritus
The recurrence relation is $a_{n+1} = \sqrt{1 - a_n}$. If you square both sides, you get:

$(a_{n+1})^2 = 1 - a_n$, or
$(a_{n+1})^2 + a_n = 1$

So $a_n$ is not squared.

Neither one. $\alpha$ is the limit of $a_n$ as $n \rightarrow \infty$. Do you know what a limit is?

We don't know the value of $\alpha$...that's what you're trying to find. But if $lim_{n \rightarrow \infty} a_n = \alpha$, then

$lim_{n \rightarrow \infty} (a_n - \alpha) = 0$

So take your recurrence relation

$(a_{n+1})^2 + a_n = 1$

and write: $a_{n+1} = \alpha + b_{n+1}$, $a_{n} = \alpha + b_n$, where $b_n$ is defined to be $a_n - \alpha$. Just plug it into the recurrence relation, and see if it tells you what must be true about $\alpha$.

7. May 9, 2017

vishnu 73

@stevendaryl
omg so sorry such a careless mistake completely did not see that !!!!

ok i have been following what you said but i lost what you mean when you when you say
so take your recurrence relation and write it as an+1 = a + bn+1 i dont how you got that from the recurrence relation please explain thanks

instead i did what bvu said
as
limn→ ∞ an+1 = an
and then using recurrence relation and quadratic formula yielded the answer please explain your method to me thanks!

8. May 9, 2017

stevendaryl

Staff Emeritus
It's not from the recurrence relation. It's just a definition. If $a_0, a_1, a_2, ...$ is any sequence converging to some constant $a$, then I can define a new sequence $b_0, b_1, ...$ by letting $b_n = a_n - a$. Then the sequence $b_0, b_1, ...$ converges to zero. That's true for any convergent sequence. For example, consider the sequence

0/1, 1/2, 2/3, 3/4, 4/5 ...

Then $a_n$ in this case would be $n/(n+1)$. The limit is 1. So we define $b_n = a_n - 1 = n/(n+1) - 1 = -1/(n+1)$. So $b_n$ is the sequence:

-1, -1/2, -1/3, -1/4, ...

which clearly converges to 0. $b_n$ is the "error" in replacing $a_n$ by its limit. The error goes to zero as $n \rightarrow \infty$

So if you rewrite the recurrence relation in terms of $a$ and $b_n$ instead of $a_n$, then it becomes:

$a_{n+1} = a + b_{n+1}$ (by definition of $b_n$)
$a_n = a + b_n$ (by definition)

So substituting gives:
$(a + b_{n+1})^2 + (a + b_n) = 1$

or rearranging to get:

$a^2 + a - 1 = - 2 a b_{n+1} - (b_{n+1})^2 - b_n$

Now, take the limit as $n \rightarrow \infty$, and you get:

$a^2 + a - 1 = 0$

9. May 9, 2017

vishnu 73

oh wow thats really smart thanks this is much more insightful solution sorry i couldn't just now see what you meant wonderful solution by the way

10. May 9, 2017

Staff: Mentor

@stevendaryl: Do you show that the series converges somewhere? If you don't do that and just assume it, setting $a_{n+1}=a_n$ and solving it for $a_n$ is easier.

11. May 9, 2017

Ray Vickson

Use a cobweb diagram, such as in http://www.milefoot.com/math/discrete/sequences/cobweb.htm

The diagram would need to be modified a bit to match the current problem, but the basic idea is the same as in the link.

12. May 9, 2017

Staff: Mentor

You can show that the sequence is contracting, or in this special case $|a_n - a| > |a_{n+1}-a|$ (the difference to the limit goes down), but that is not trivial.

Consider a sequence $s_{n+1}= 2 s_n - 1$.
If we know $s_0=1$ it leads to $s_n=1$ for all $n$ and this sequence is convergent. But every other starting value leads to a diverging series.

13. May 9, 2017

stevendaryl

Staff Emeritus
Yeah, it amounts to the same thing. I thought it might be confusing to do that, because, of course, $a_{n+1} \neq a_n$ for any $n$.

It just occurred to me that you can take the recurrence equation,

$(a_{n+1})^2 = 1 - a_n$

and just take the limit of both sides.

14. May 9, 2017

Ray Vickson

If you pay attention to post #11 you will see that not only does the limit $\lim_n a_n = a$ exist, but also that it equals the value where the two graphs $y = a$ and $y = \sqrt{1-a}$ intersect.

15. May 9, 2017

bobstar

Search how to get the formula for the golden ratio Fibonacci series. Simple!

16. May 10, 2017

vishnu 73

oh wow all of you all are really smart having so many ingenious methods really smart ency you all sorry i am just starting on olympiads please dont mind my mathematical immaturity
once thanks to all