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Find the value of
$$\lim_{n\to \infty}\bigg(1+\dfrac{1}{n}\bigg)\bigg(1+\dfrac{2}{n}\bigg)^{\frac12}\ldots(2)^{\frac{1}{n}}$$

My work:
$\bigg(1+\dfrac{1}{n}\bigg)=\bigg\{\bigg(1+\dfrac{1}{n}\bigg)^n\bigg\}^{\frac{1}{n}}=e^{\frac{1}{n}}$
$\bigg(1+\dfrac{2}{n}\bigg)^{\frac12}=\bigg\{\bigg(1+\dfrac{2}{n}\bigg)^{\frac{n}{2}}\bigg\}^{\frac{1}{n}}=e^{2\cdot\frac12\cdot\frac{1}{n}}=e^\frac{1}{n}$
$~~~~~~~~~~~~\vdots$
$~~~~~~~~~~~~\vdots$
$\bigg(1+\dfrac{n}{n}\bigg)^{\frac{1}{n}}=e^{\frac{1}{n}}$
So, $L=e$
But, the answer says $L=e^{\frac{\pi^2}{12}}$.
I do not know where I am going wrong, is the answer a typo or I am doing wrong. Please help.

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  • $\begingroup$ Have you considered taking the log and showing it tends to $\dfrac{\pi^2}{12}$? $\endgroup$ – user88595 Feb 20 '14 at 13:39
  • $\begingroup$ No, I have not, but why should I do that, if there is no mistake in the technique I am using? $\endgroup$ – Hawk Feb 20 '14 at 13:41
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    $\begingroup$ How come $\left(1+\frac1n\right)\doteq e^{\frac1n}$? $\endgroup$ – Ruslan Feb 20 '14 at 13:46
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    $\begingroup$ Numerically I check that the limit is $2.276108 \approx e^{\pi^2/12}$. You are assuming too much using identities as in you use $=$ when you should be using $\approx$. $\endgroup$ – user88595 Feb 20 '14 at 13:53
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    $\begingroup$ Firstly $x_n \approx y_n$, does not imply $\prod_{n=1}^{\infty} x_n = \prod_{n=1}^{\infty} y_n$, for example $(1+1/n) \approx 1$, but $\lim (1+ 1/n)^n$ is not $1$. $\endgroup$ – r9m Feb 20 '14 at 14:02
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This seems to be the reasoning in your argument $$ \begin{align} \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac kn\right)^{1/k} &=\lim_{n\to\infty}\left(\prod_{k=1}^n\left(1+\frac kn\right)^{n/k}\right)^{1/n}\tag{1}\\ &=\lim_{n\to\infty}\left(\prod_{k=1}^n\lim_{n\to\infty}\left[\left(1+\frac kn\right)^{n/k}\right]\right)^{1/n}\tag{2}\\ &=\lim_{n\to\infty}\left(\prod_{k=1}^n\ e\right)^{1/n}\tag{3}\\[12pt] &=\ e\tag{4} \end{align} $$ All of the steps are fine except $(2)$. It is not, in general, allowed to take the limit of an inner part like that. For example consider $$ \begin{align} \lim_{n\to\infty}\left(\frac1n\cdot n\right) &=\lim_{n\to\infty}\left(\lim_{n\to\infty}\left[\frac1n\right] n\right)\tag{5}\\ &=\lim_{n\to\infty}\left(0\cdot n\right)\tag{6}\\[3pt] &=\lim_{n\to\infty}\ 0\tag{7}\\[2pt] &=\ 0\tag{8} \end{align} $$ Step $(5)$ is the same as step $(2)$, but that step allows us to show that $1=0$.

To see why this affects your limit adversely, notice that no matter how big $n$ gets in the limit, when $k$ is near $n$, $\left(1+\frac kn\right)^{n/k}$ is close to $2$, not $e$. Thus, the terms of the product are between $2$ and $e$. Not all of them tend to $e$.


What we need to do is use the continuity of $\log(x)$ as viplov_jain suggests. $$ \begin{align} \log\left(\lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac kn\right)^{1/k}\right) &=\lim_{n\to\infty}\log\left(\prod_{k=1}^n\left(1+\frac kn\right)^{1/k}\right)\tag{9}\\ &=\lim_{n\to\infty}\sum_{k=1}^n\frac1k\log\left(1+\frac kn\right)\tag{10}\\ &=\lim_{n\to\infty}\sum_{k=1}^n\frac nk\log\left(1+\frac kn\right)\frac1n\tag{11}\\ &=\int_0^1\frac1x\log(1+x)\,\mathrm{d}x\tag{12}\\ &=\int_0^1\sum_{k=0}^\infty(-1)^k\frac{x^k}{k+1}\,\mathrm{d}x\tag{13}\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^2}\tag{14}\\ &=\frac{\pi^2}{12}\tag{15} \end{align} $$ Step $(12)$ uses the idea of approximating a Riemann Sum by an integral. $(15)$ tells us that $$ \lim_{n\to\infty}\prod_{k=1}^n\left(1+\frac kn\right)^{1/k}=e^{\pi^2/12}\tag{16} $$ Notice that $$ 2\lt2.27610815162573\doteq e^{\pi^2/12}\lt e\tag{17} $$

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  • $\begingroup$ Awesome solution as always. Thank you. $\endgroup$ – Hawk Feb 27 '14 at 11:36
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Let $x$ be the value of the limit .Take $\log(x)$ to get $$\lim_{n\to \infty}\sum_{i=1}^n \frac{1}{i/n}\log(1+i/n)\cdot\frac{1}{n}$$ which is equal to $\int_0^1 \frac{\log(1+x)}{x}\mathrm{d}x$ which is $\pi ^2/12.$

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  • $\begingroup$ This is what @user88595 said, I do not have any objection to that, I want to know why should I take $\log$, is there any error in my method? $\endgroup$ – Hawk Feb 20 '14 at 13:53
  • $\begingroup$ @Hawk the error in your method is that you can not write $(2)^{1/n}=e^{1/n}$, you need to be careful for this kind of sequence where the number of terms in the product for the $n$ th term of sequence is itself changing. $\endgroup$ – Samrat Mukhopadhyay Feb 20 '14 at 13:55
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    $\begingroup$ Also writing $$\left(\left(1+\frac1n\right)^n\right)^n=e^{1/n}\;$$ is a serious mistake. $\endgroup$ – DonAntonio Feb 20 '14 at 13:56
  • $\begingroup$ Calculating indeterminate forms term wise is applicable for finite sums or products only. $\endgroup$ – viplov_jain Feb 20 '14 at 13:56
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    $\begingroup$ @DonAntonio I understand now...thanks... $\endgroup$ – Hawk Feb 20 '14 at 13:57

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