# 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

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

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

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

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

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