Can i do this to an infinite series?

Question

$$let , Y=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}$$ $$Then , Y= \sqrt{2Y}$$ $$Y^2 = 2Y$$ $$Y^2 - 2Y = 0$$ $$Y = 0 , Y =2$$ Now y can't be zero so , $$Y = 2$$ Is This Correct?

So far you prove: If it converges, then it converges to $2$. The other part is to show it converges. — GEdgar, Aug 27 '19 at 21:45
You have a sequence here, defined recursively as $y_0 = \sqrt{2}$ and $y_{n+1} = \sqrt{2y_n}$. Use induction to prove $y_n$ is increasing and bounded -- hence, convergent. — RRL, Aug 27 '19 at 21:50

score 2 · Accepted Answer · answered Aug 27 '19 at 22:42

If you assume the number exists and has a value then that is true.

But it could be that the number doesn't exist.

Consider $1 + 2 + 4 + 8 + 16+..... = M$ then $2M = 2+4 + 8 + 32 + ....$ and $2M + 1 = 1 + 2 + 4+ 8+16 + 32 + .... = M$ so $2M + 1 = M$ and $M = -1$.

That's obviously not true.

ANd this fails because $\lim\limits_{n\to \infty}\sum\limits_{k=0}^n 2^k=\infty$.

However If $a_0 =\sqrt{2}$ and $a_1= \sqrt {2\sqrt 2}$ and $a_{k+1} =\sqrt {2a_k}$ then IF $\{a_n\}$ converges so $Y = \lim\limits_{n\to \infty} a_n\ne \infty$, we would be okay. And we could say $Y=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}$ and $Y= 2$ by your argument.

We can prove $\lim\limits_{n\to \infty} a_n\ne \infty$ by noting that $a_0=\sqrt 2 < 2$ and if $a_k <2$ we would have $a_{k+1} =\sqrt {2a_k}< \sqrt {2*2} = 2$. so by induction all $a_n < 2$ so $\lim\limits_{n\to \infty} a_n\ne \infty$.

Furthermore $a_{k+1} = \sqrt{2a_k} > \sqrt{a_k*a_k} = a_k$.

So we have $\sqrt{2} =a_1 < a_2 < a_3 < ....... < 2$. So $\lim\limits_{n\to \infty} a_n = Y$ exists.

score 0 · Answer 2 · answered Aug 27 '19 at 21:42

0

You misplaced the minus sign. Indeed, the given equation reduces to \begin{align*} Y = \sqrt{2Y} \Longleftrightarrow Y^{2} = 2Y\,\,\wedge\,\,Y \geq 0 \Longleftrightarrow Y = 0\,\,\vee\,\, Y = 2 \end{align*}

Since $Y > 0$, the sought result is given by $Y = 2$.

answered Aug 27 '19 at 21:42

user0102

21,867

How do you know that $Y>0$? – Axion004 Aug 27 '19 at 21:43
1

You can also think of it as \begin{align} 2^{1/2 + 1/4 + \ldots} = 2^{1} = 2 \end{align}
That is why.
– user0102 Aug 27 '19 at 21:48

score 0 · Answer 3 · answered Aug 27 '19 at 22:18

You are assuming that the sequence is convergent and found the limit to be $2$

You need to prove that the sequence is indeed convergent.

The proof is based on the fact that your sequence is increasing and it is bounded above by $M=2$ So go ahead and finish the proof.

Can i do this to an infinite series?

3 Answers3