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Evaluate: $3\cdot9^{\frac{1}{2}}\cdot27^{\frac{1}{4}}\cdot81^{\frac{1}{8}} \dotsb$

Trial: Let $$\begin{align} P &= 3 \cdot 9^{\frac{1}{2}} \cdot 27^{\frac{1}{4}} \cdot 81^{\frac{1}{8}} \dotsb\\ \implies \ln P &=\ln3+\frac{1}{2} \ln 3^2+\frac{1}{2^2} \ln 3^3+\frac{1}{2^3} \ln 3^4 + \dotsb\\ &=\ln 3 \sum_{x=0}^{\infty}\frac{x+1}{2^x} \end{align}$$

Then how I proceed. Is there any other simpler way to solve. Please help.

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Is what you wish to evaluate the infinite product, or just the first four terms? –  Herng Yi Jan 20 '13 at 12:31
    
@HerngYi: Infinite product –  A.D Jan 20 '13 at 12:32
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4 Answers 4

up vote 1 down vote accepted

If $$S_n=\sum_{0\le r\le n}(r+1)u^r=1+2u+3u^2+\cdots+nu^{n-1}+(n+1)u^n$$

So, $$uS_n=u+2u^2+3u^3+\cdots+nu^{n}+(n+1)u^{n+1}$$

So, $$S_n-uS_n$$ $$=1+u(2-1)+u^2(3-2)+\cdots+u^{n-1}\{n-(n-1)\}+u^n\{(n+1)-n\}-(n+1)u^{n+1}$$ $$=1+u+u^2+\cdots+u^{n-1}+u^n-(n+1)u^{n+1}$$ $$=\frac{1-u^{n+1}}{1-u}-(n+1)u^{n+1}$$

If $|u|<1, \lim_{n\to \infty} u=0$ and $\lim_{n\to \infty} n u^n=0\text{ (Proof below)}$ then $$\lim_{n\to \infty}(1-u)S_n=\frac1{1-u}\iff \lim_{n\to \infty}S_n=\frac1{(1-u)^2}$$

So putting $u=\frac12$, $$\lim_{n\to \infty}\sum_{0\le r\le n}(r+1)\left(\frac12\right)^r=\frac1{\left(\frac12\right)^2}=4$$

Hence, $\log P=4\log 3\iff P=3^4=81$

[

Proof:

$\lim_{n\to \infty} n u^n=\lim_{n\to \infty}\frac{n}{\left(\frac1u\right)^n}$ ($\frac \infty\infty$ form)

Applying L'Hospital’s Rule,

$\lim_{n\to \infty} n u^n=\lim_{n\to \infty} \frac1{\left(\frac1u\right)^{n-1}\ln(\frac1u)}=-\lim_{n\to \infty}\frac{u^{n-1}}{\ln u}=0$ if $|u|<1$

It can also be proved using Pringsheim's theorem.

]

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$$f(x):=\sum_{k=0}^\infty x^k=\frac{1}{1-x}\,\,,\,\,|x|<1\Longrightarrow f'(x)=\frac{1}{(1-x)^2}=\sum_{k=1}^\infty kx^{k-1}\,\,,\,\,|x|<1$$

But then

$$\sum_{k=0}^\infty\frac{k+1}{2^k}=\sum_{k=1}^\infty k\left(\frac{1}{2}\right)^k+\sum_{k=0}^\infty\left(\frac{1}{2}\right)^k=\ldots$$

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I think you are correct so far. Use the facts that

$$\sum_{k=0}^{\infty} r^k = \frac{1}{1-r} $$

and

$$\sum_{k=0}^{\infty} k r^k = \frac{r}{(1-r)^2} $$

So you will have, with $r=1/2$:

$$\ln{P} = (2 + 2) \ln{3} = 4 \ln{3} \implies P = 81 $$

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Sorry, but, Is not every single term greater than one. So should not the product of infinite terms greater than one be infinite? Logically, what is it that makes it converge? I hope you will not mind to answer that. –  007resu Jan 20 '13 at 13:09
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No, I should have specified that $|r| < 1$, which is the case here. Powers increase far faster than linear terms, so the sum converges. For products, it is only important that the sequence converges to one from above sufficiently fast, which is the case with your product. –  Ron Gordon Jan 20 '13 at 13:15
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We have:

$$ P = \prod_{k=1}^\infty \big(3^k\big)^{\left(2^{1-k}\right)}$$

Take the $\log_3$ of both sides:

$$ \log_3 P = \sum_{k=1}^\infty \log_3 3^{k2^{1-k}} $$ $$ \log_3 P = \sum_{k=1}^\infty k2^{1-k} $$ $$ \log_3 P = 2\sum_{k=1}^\infty k\left(\frac{1}{2}\right)^k$$

It is known that: $$ \sum_{k=1}^{\infty} k r^k = \frac{r}{\left(1-r\right)^2} $$

So:

$$ \log_3 P = 2 \cdot \frac{\frac{1}{2}}{(1-1/2)^2} $$ $$ \log_3 P = \frac{1}{1/4} $$ $$ \log_3 P = 4 $$ $$ P = 3^4 = 81 $$

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