17
$\begingroup$

Evaluate: $$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots \cdot\sqrt[2^n]{2^n}$$

My attempt:First solve when $n$ is not infinity then put infinity in.

$$2^{\frac{1}{2}}\cdot 4^{\frac{1}{4}}\cdot \dots\cdot (2^n)^{\frac{1}{2^n}}$$

$$=2^{\frac{1}{2}}\cdot 2^{\frac{2}{4}}\cdot \dots\cdot 2^{\frac{n}{2^n}}$$

Now calculate the sum of the powers:

$$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\dots+\frac{n}{2^n}$$

$$=\frac{2^{n-1}+2\cdot2^{n-2}+3\cdot2^{n-3}+\dots+n\cdot2^0}{2^n}$$

Now calculate the numerator:

$$2^0+2^1+2^2+\dots+2^{n-1}=2^n-1$$

$$+$$

$$2^0+2^1+\dots+2^{n-2}=2^{n-1}-1$$

$$+$$

$$2^0+2^1+\dots+2^{n-3}=2^{n-2}-1$$

$$+$$

$$\vdots$$

$$+$$

$$2^0=2^1-1$$

$$=2^1+2^2+2^3+\dots+2^n-n=2^{n+1}-n-1$$

Now put the numerator on the fraction:

$$\frac{2^{n+1}-n-1}{2^n}=2-\frac{n}{2^n}-\frac{1}{2^n}$$

Now we can easily find $\lim_{n \to \infty}\frac{1}{2^n}=0$

Then we just have to find $\lim_{n \to \infty }\frac{n}{2^n}$, that by graphing will easily give us the answer zero.

That gives the total answer is $4$.

But now they are two problems:

1.I cannot find $\lim_{n \to \infty }\frac{n}{2^n}$ without graghing.

2.My answer is too long.

Now I want you to help me with these problems.Thanks.

$\endgroup$
6
  • 1
    $\begingroup$ Hint: use that $\log(ab)=\log(a)+\log(b)$ and convert it into a summation (series) problem. $\endgroup$ Commented Aug 5, 2016 at 20:10
  • $\begingroup$ @Fimpellizieri We didn't learn logarithem yet. $\endgroup$ Commented Aug 5, 2016 at 20:11
  • 1
    $\begingroup$ Idk why no one has mentioned this yet, but $\sum \frac{n}{2^n}$ looks an awful lot the series expansion of $\left(\frac{1}{1-x}\right)^2$, it's just missing a factor of $x$, that is $x\left(\frac{1}{1-x}\right)^2$. So $\sum \frac{n}{2^n}=\frac{1}{2}\frac{1}{1/4}=2$, since $|1/2|<1$. $\endgroup$
    – Nobody
    Commented Aug 5, 2016 at 20:52
  • $\begingroup$ thanks for showing your work, +1 $\endgroup$
    – jimjim
    Commented Aug 6, 2016 at 1:43
  • $\begingroup$ It seems weird to have limits introduced but not logarithms. $\endgroup$
    – Batman
    Commented Aug 6, 2016 at 1:52

7 Answers 7

29
$\begingroup$

$$I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}+\cdots$$ $$2I=1+1+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\frac{6}{32}+\cdots$$ $$2I-I=1+\left(1-\frac 12 \right)+\left(\frac 34 -\frac 24 \right)+\left(\frac 48 -\frac 38 \right)+\left(\frac {5}{16} -\frac {4}{16} \right)+\cdots$$ $$I=1+\frac 12+\frac 14+\frac 18+\cdots=2$$ therefore $$\lim_{n\to \infty }\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots\times\sqrt[2^n]{2^n}=2^2=4$$

$\endgroup$
5
  • 3
    $\begingroup$ Nice answer it is too much short and easy to under stand +1. $\endgroup$ Commented Aug 5, 2016 at 20:13
  • 1
    $\begingroup$ Would you be so kind as to explain the "therefore"? I'm having trouble seeing the jump which apparently is obvious. O_o $\endgroup$
    – John
    Commented Aug 5, 2016 at 21:28
  • 6
    $\begingroup$ @John $\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac 12+\frac 24+\frac 38+\cdots}=2^2=4$ $\endgroup$ Commented Aug 5, 2016 at 21:30
  • 4
    $\begingroup$ Incredibly clever! $\endgroup$
    – John
    Commented Aug 5, 2016 at 21:32
  • $\begingroup$ @John Please. So thanks. $\endgroup$ Commented Aug 5, 2016 at 21:33
11
$\begingroup$

I believe this is a "known" representation of the key summation:

enter image description here

$$\sum_{n=1}^{\infty} \frac{n}{2^n} \;=\; \frac{1}{2} + \left( \frac{1}{4} + \frac{1}{4} \right) + \left( \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) + \cdots \;=\; 2 $$

$\endgroup$
0
4
$\begingroup$

I hope this would be useful for you.

So your trying to evaluate $\displaystyle\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}$. Consider first $\displaystyle\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}$, after developing the product properly. Now you have to make sense of $$\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}$$ , so to reach a specific value, note that $\sum_{n=1}^{k}\frac{n}{2^{n}}=\sum_{n=1}^{k}\frac{n-1}{2^{n-1}}-\frac{n}{2^{n}}+\sum_{n=1}^{k}\frac{1}{2^{n-1}}$, therefore

\begin{eqnarray*} \lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}&=&\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n-1}{2^{n-1}}-\frac{n} {2^{n}}+\sum_{n=1}^{k}\frac{1}{2^{n-1}}\\ &=&\lim_{k\to\infty}-\frac{k}{2^{k}}+\frac{1-\frac{1}{2^{k+1}}}{1-\frac{1}{2}}\\ &=&2 \end{eqnarray*} Finally $$\prod_{n=1}^{\infty}\sqrt[2^{n}]{2^{n}}=\lim_{k\to\infty}\prod_{n=1}^{k}\sqrt[2^{n}]{2^{n}}=\lim_{k\to\infty}2^{\sum_{n=1}^{k}\frac{n}{2^{n}}}=2^{\lim_{k\to\infty}\sum_{n=1}^{k}\frac{n}{2^{n}}}=2^{2}=4$$

$\endgroup$
2
  • $\begingroup$ Even though, it solves the problem, it doesn't answer your question, sorry for that. $\endgroup$ Commented Aug 5, 2016 at 20:48
  • $\begingroup$ I don't really want the answer of question I wanted a shorter answer.+1 $\endgroup$ Commented Aug 6, 2016 at 5:07
2
$\begingroup$

$$\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots$$

When we use the LOG function get (when $n$ is positive):

$$\ln\left(\left(2^n\right)^{\frac{1}{2^n}}\right)=\frac{n\ln(2)}{2^n}$$

And using, when $a$ and $b$ are positive:

$$\ln(ab)=\ln(a)+\ln(b)$$

So, we get that:

$$\ln(\text{P})=\frac{1\ln(2)}{2^1}+\frac{2\ln(2)}{2^2}+\frac{3\ln(2)}{2^3}+\dots=\ln(2)\left[\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\dots\right]$$

And, now we know that:

$$\sum_{n=1}^{\infty}\frac{n}{2^n}=\frac{1}{2^1}+\frac{2}{2^2}+\frac{3}{2^3}+\dots=\frac{2}{(2-1)^2}=2$$

Using, when $|x|>1$:

$$\sum_{n=1}^{\infty}\frac{n}{x^n}=\frac{x}{(x-1)^2}$$

So:

$$\ln(\text{P})=\ln(2)\left[2\right]=2\ln(2)\Longleftrightarrow\text{P}=4$$

We, find the answer:

$$\color{red}{\text{P}=\prod_{n=1}^{\infty}\sqrt[2^n]{2^n}=\sqrt[2]{2}\times\sqrt[4]{4}\times\sqrt[8]{8}\times\dots=2^{\frac{1}{2}}\times4^{\frac{1}{4}}\times8^{\frac{1}{8}}\times\dots=4}$$

$\endgroup$
3
  • $\begingroup$ OP says they did not learn logarithms yet and anyway logarithms are not useful to solve these, only the rule $a^xa^y=a^{x+y}$ is. Are these the reasons why you focus on logarithms? $\endgroup$
    – Did
    Commented Aug 10, 2016 at 9:50
  • $\begingroup$ Not usefull do not mean, wrong! And there was an answer that have a more easy way, what I did is just another way! $\endgroup$ Commented Aug 10, 2016 at 9:54
  • $\begingroup$ Right, "not useful" means "not useful". By the way, do you seriously think that drawing the identity $\sum\limits_{n=1}^{\infty}\frac{n}{x^n}=\frac{x}{(x-1)^2}$ out of the hat makes for a good post (i) for the OP, (ii) for other users? The intersection of the populations of users who need an answer and who understand this step might be empty... $\endgroup$
    – Did
    Commented Aug 10, 2016 at 9:57
1
$\begingroup$

$I=1+\frac 12+\frac 14+\frac 18+\frac 1{16}+\cdots=2$

$I^2 =1+\frac{2}{2}+\frac{3}{4}+\frac{4}{8}+\frac{5}{16}+\cdots$

$I^2 -I=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+\frac{4}{16}+\cdots=2^2-2=2$

Then

$\lim_{n\to \infty }\sqrt[2]{2}\cdot \sqrt[4]{4}\cdot \sqrt[8]{8}\cdot \dots\cdot \sqrt[2^n]{2^n}=2^{I^2-I}=2^2=4$

$\endgroup$
1
  • 2
    $\begingroup$ The second line might require some explanation but I got it +1 ;) $\endgroup$ Commented Aug 6, 2016 at 9:07
1
$\begingroup$

A standard thing to do to, is too use the commonly known geometric series formula, and set:

$$f(x)=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}$$

The formula is not hard to prove, and it holds for $|x|<1$ (which can be seen by the ratio test for convergence) we take $x \in (-1,1)$.

Then by term by term differentiation we get:

$$f'(x)=\sum_{n=1}^{\infty} nx^{n-1}=\frac{1}{x}\sum_{n=1}^{\infty} nx^n=\frac{1}{(1-x)^2}$$

Hence we have:

$$\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$$

And the case your interested about is $x=\frac{1}{2}$

$\endgroup$
0
$\begingroup$

I would like to generalize Behrouz' rather nifty answer:

$$I=\sum_{n=1}^{\infty} \frac{n}{r^n}=\frac{1}{r}+\frac{2}{r^2}+\frac{3}{r^3}+\frac{4}{r^4}$$ The key is now to multiply everything by $r$:

$$rI=1+\frac{2}{r}+\frac{3}{r^2}+\frac{4}{r^3}$$ Now we calculate $rI-I$ $$rI-I=1+(\frac{2}{r}-\frac{1}{r})+(\frac{3}{r^2}-\frac{2}{r^2})+(\frac{4}{r^3}-\frac{3}{r^3})+...$$ simplifies to

$$1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}...$$ And that is our good old geometric sum

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .