Is this correct?

**Henryt999** · February 13th 2010, 08:36 AM

Are these answers I came up with acctually solutions or more just idéas that doesn´t acctually show anything and by "coincidence" happend to have the same numbers as the actual solutions?
Question is:
Look at $[a_n]_{n=1}^{\infty}$
It is defined by: $a_{n+1} = a_n-a_n^2$ with $a_0$ = $\frac{1}{2}$ and $0\leq n$

a) Find the limit of the recursive formula.
b)Find the limit of $\lim_{n\rightarrow \infty} \sum_{k=0}^{n}a_k^2$ this one I´m not sure I understand everywhere is $a_n$ suddenly we have $a_k$ and then from k=0 where do I put the 0. Or is it just that the sum from.. $a_1$ ..oh well lets just assume it means from $a_1$

a: Solution: Since the limit of $a_{n+1}$ =limit $a_n$
we can write (I think) $a=a-a^2$
this is only satisfied if a = 0. Hence the limit must be 0....hmm book said 0 too, well then something must be right (maybe)

b) Solution: $\lim_{n\rightarrow \infty} \sum_{k=0}^n a_k^2 = \lim_{n\rightarrow\infty}(\frac{1}{2}-(\frac{1}{n})^n) = \lim_{n\rightarrow \infty} (\frac{1}{2}-\frac{1^n}{n^n}) = \frac{1}{2}$ . Book say 1/2..Wow I must have done something correct, ooor there was a $\frac{1}{2}$ that magicaly appered

**mabruka** · February 13th 2010, 09:59 AM

Remember the little letters are just index character and do not have a special meaning beyond that. You can use whatever you want.

if k=1,2,... and n=1,2,... then $(a_k)_k=(a_n)_n$

With that said yhou are asked to compute

$<br /> \lim_{n\rightarrow \infty} \sum_{k=0}^{n}a_k^2$ dont sweat about it being changed

So yo have that $\lim_{n\rightarrow \infty} \sum_{k=0}^{n}a_k^2 = \lim_{k\rightarrow \infty} \sum_{n=0}^{k}a_n^2$

I would suggest that you use the recursion formula you know to get a telescopic series and see what happens

**mabruka** · February 13th 2010, 10:02 AM

Oh and about a) i think you did right.

**Defunkt** · February 13th 2010, 03:06 PM

Originally Posted by Henryt999

Are these answers I came up with acctually solutions or more just idéas that doesn´t acctually show anything and by "coincidence" happend to have the same numbers as the actual solutions?
Question is:
Look at $[a_n]_{n=1}^{\infty}$
It is defined by: $a_{n+1} = a_n-a_n^2$ with $a_0$ = $\frac{1}{2}$ and $0\leq n$

a) Find the limit of the recursive formula.
b)Find the limit of $\lim_{n\rightarrow \infty} \sum_{k=0}^{n}a_k^2$ this one I´m not sure I understand everywhere is $a_n$ suddenly we have $a_k$ and then from k=0 where do I put the 0. Or is it just that the sum from.. $a_1$ ..oh well lets just assume it means from $a_1$

a: Solution: Since the limit of $a_{n+1}$ =limit $a_n$
we can write (I think) $a=a-a^2$
this is only satisfied if a = 0. Hence the limit must be 0....hmm book said 0 too, well then something must be right (maybe)

b) Solution: $\lim_{n\rightarrow \infty} \sum_{k=0}^n a_k^2 = \lim_{n\rightarrow\infty}(\frac{1}{2}-(\frac{1}{n})^n) = \lim_{n\rightarrow \infty} (\frac{1}{2}-\frac{1^n}{n^n}) = \frac{1}{2}$ . Book say 1/2..Wow I must have done something correct, ooor there was a $\frac{1}{2}$ that magicaly appered

For the first question, you can only use arithmetic of limits if you proved that the sequence converges, which you did not.

To prove the sequence converges, show that it is bounded and monotonically decreasing. Only after you show that you can proceed to the final step which you described.

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