Find $\displaystyle\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{1}{2n}\right)\left(1+\dfrac{1}{4n}\right)\ldots\left(1+\dfrac{1}{2^{n-1}n}\right)$.
(This is not my homework. One of my friends gave this to me.)
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Apply the GM-AM inquality to get:
$$\begin{align} \prod_{j=1}^n\left(1+\frac1{2^{j-1}n}\right)&\leq\left(1+\frac1{n^2}\frac{1+\cdots+2^{n-1}}{2^{n-1}}\right)^n\\ &\leq\left(1+\frac2{n^2}\right)^n \end{align}$$
Since last expression tends to $1$, the limit is $\leq1$
Since each factor is greater than $1$, the limit must be $1$.
ajotatxe
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1what's a DM inequality? – Alex Jun 23 '14 at 11:59
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Sorry for the typos ant mistakes. I think that now it is correct. – ajotatxe Jun 23 '14 at 12:10
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1How do you apply the AM-GM inequality ? I suppose you use $(x_1x_2\ldots x_n)^{\frac{1}{n}} \leq \frac{x_1+x_2+\ldots+x_n}{n}$ for some $x_1,x_2,\ldots,x_n$. Can you tell me which formula you use for $x_i$ ? – Ewan Delanoy Jun 23 '14 at 12:12
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$x_j=1+\frac1{2^{j-1}n}$ – ajotatxe Jun 23 '14 at 12:20
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Correct me if I'm wrong, but you only show the upper bound here. You also need to show the lower bound to establish the limit of the original sequence. – Alex Jun 23 '14 at 13:06
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1@Alex: "Since each factor is greater than 1, the limit must be 1." – ajotatxe Jun 23 '14 at 13:08
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Let $y$ be the value of limit. Then, $$\ln y=\lim_{n\rightarrow \infty} \sum_{k=1}^{n} \ln\left(1+\frac{1}{2^{k-1}n}\right)$$ Since the denominator in $\dfrac{1}{2^{k-1}n}$ is very large, we can work under the approximation $$\ln\left(1+\frac{1}{2^{k-1}n}\right) \approx \frac{1}{2^{k-1}n}$$ Hence, $$\ln y=\lim_{n\rightarrow \infty} \sum_{k=1}^n \frac{1}{2^{k-1}n}=\lim_{n\rightarrow \infty} \frac{2}{n}\left(1-\frac{1}{2^n}\right)=0$$ $$\Rightarrow y=e^0=\boxed{1}$$
Pranav Arora
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I don't find that not bounded approximations and "large numbers" can be in a proof. – ajotatxe Jun 23 '14 at 12:22
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@ajotatxe: I myself think that the solution is incorrect. It looks as if I posted a similar answer to Mehdi stating it in a different way. – Pranav Arora Jun 23 '14 at 12:24
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If you put $\leq$ instead of $\approx$ you get an upper bound for the limit, and the lower bound is easy. – ajotatxe Jun 23 '14 at 12:28
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@ajotatxe: I am sorry if I am being dumb but is the lower bound zero? – Pranav Arora Jun 23 '14 at 12:31
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Yes, the point is that upper and lower bound are the same, so you can say what exactly the limit is. – ajotatxe Jun 23 '14 at 12:42
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Isn't it $1$? The fractional expressions inside the parentheses are zero as $n\to \infty$.
$$\lim\limits_{n\to\infty}\frac{1}{n}=0$$ $$\lim\limits_{n\to\infty}\frac{1}{2n}=0$$ $$\vdots$$ $$\lim\limits_{n\to\infty}\frac{1}{2^{n-1}n}=0$$
Consequently the answer is $1$
M.X
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3You have to be careful, because the number of factors doubles when you increase $n$ by $1$. So even though the factors reduce in size it is not so immediate that this is the dominant feature of the product. – Mark Bennet Jun 23 '14 at 12:02