How to calculate the following limit $$ \lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots \left(1+\frac{n}{n}\right) \right]^\frac{1}{n} .$$ I was trying this by taking the $\log $ of the product and then limit but I am not getting the answer could anybody please help me. And, also is there any general rule for calculating the limit for $$ (a_1a_2.\dots a_n)^{\frac{1}{n}}.$$
Thanks.
On
Hint:
$$S=\lim _{ n\rightarrow \infty } \left[ \left( 1+\frac { 1 }{ n } \right) \left( 1+\frac { 2 }{ n } \right) \cdots \left( 1+\frac { n }{ n } \right) \right] ^{ \frac { 1 }{ n } }\\ \ln { S } =\lim _{ n\rightarrow \infty } \frac { 1 }{ n } \sum _{ k=1 }^{ n }{ \ln { \left( 1+\frac { k }{ n } \right) } } =\int _{ 0 }^{ 1 }{ \ln { \left( 1+x \right) dx } } $$
On
Taking the $\log$ of the expression, we arrive to the sum
$$ \frac{1}{n} \sum_{k=1}^n \log \left( 1 + \frac{k}{n} \right). $$
This sum is a Riemann sum for $\int_1^2 \log(x) \, dx = \left[ x\log x - x\right]_{x=1}^{x=2} = 2\log(2) - 1$. Hence, the limit is
$$ e^{2\log(2) - 1} = \frac{4}{e}.$$
On
You should recognize the logarithm as
$$\frac{1}{n} \sum_{k = 1}^n \ln\left(1 + \frac k n\right)$$
which is a Riemann sum for $\int_1^2 \ln x \, dx$.
As far as general rules, things are frequently a bit ad hoc. If you can find a Riemann sum, it's helpful. That's frequently going to happen precisely because of the factor $1/n$ that you can pull down with a log.
On
The log of the product: $$\frac1n\sum_{k=1}^n\ln\Bigl(1+\frac kn\Bigr)$$ is a Riemann sum for the function $\ln(1+x)$, between the values $x=0$ and $x=1$, hence the limit is $$\int_0^1\ln(1+x)\,\mathrm d\mkern1mu x =\int_1^2\ln u\,\mathrm d\mkern1mu u=u\ln u-u\,\bigg\vert_{u=1}^{u=2} =2\ln 2-1,$$ hence the limit is $\;\mathrm e^{2\ln 2-1}=\dfrac 4{\mathrm e}.$
On
Answering your second question as the previous answers all answer your first.
Select a sequence $\{b_n\}$ such that $a_n = e^{b_n}$
Now, instead, we have ${\left({e^{\left(\sum\limits_{i=1}^n b_i \right)}}\right)^{\frac 1 n}} = \left(\prod\limits_{i=1}^n a_i \right)^{\frac 1 n}$
This may allow you to leverage identities involving infinite sums as opposed to infinite products as $n \to \infty$. However, you can't always expect certain properties of the sequence to be preserved by mapping them as powers of $e$.
On
What you are looking for in general is called a geometric integral:
$$\prod_a^b f(x)^{dx} = \lim_{\Delta x \to 0}\prod_{i} f(x_i)^{\Delta x}$$
Where the $x_i$ range from $a$ to $b$ with difference $\Delta x$. This can be seen as a multiplicative analog of the Riemann integral. In fact, it can be shown that in general:
$$\prod_a^b f(x)^{dx} = \exp\left(\int_a^b\ln\left(f(x)\right)\,dx\right)$$
by simply writing out the Rienmann sum for $\ln\left(f(x)\right)$. In this particular case, we get:
$$\lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots \left(1+\frac{n}{n}\right) \right]^\frac{1}{n} = \exp\left(\int_0^1\ln(1+x)\,dx\right) = \frac4e$$
On
This answer uses Sterling's approximation to get the limit:
\begin{align} l=\;&\left[ \left( 1+\frac { 1 }{ n } \right) \left( 1+\frac { 2 }{ n } \right) \cdots \left( 1+\frac { n }{ n } \right) \right]^{ \frac { 1 }{ n } } \\ =\; & \left[ \frac{1}{n^n}(n+1)(n+2)\dots(n+n) \right]^{ \frac { 1 }{ n } } \\ = \;& \frac{1}{n}\left[ \frac{(2n)!}{n!} \right]^{ \frac { 1 }{ n } } \\ \approx\; & \frac{1}{n}\left[\sqrt{2}\cdot 2^{2n}\left(\frac{n}{e} \right)^n\right]^{\frac{1}{n}} \Rightarrow \\ \lim_{n\rightarrow \infty}l= \;& \frac{4}{e} \end{align}
On
$$\lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots \left(1+\frac{n}{n}\right)\right]^\frac{1}{n}$$
$$=\lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)^n\left(1+\frac{2}{n}\right)^n\cdots \left(1+\frac{n}{n}\right)^n\right]^\frac{1}{n^2}$$
Now use the characteristic of $e$ that $\lim_{n \to \infty}(1+\frac m n)^{n}=\lim_{n \to \infty}(1+\frac 1 n)^{m \,n}$
$$=\lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)^n\left(1+\frac{1}{n}\right)^{2n}\cdots \left(1+\frac{1}{n}\right)^{n^2}\right]^\frac{1}{n^2}$$
$$=\lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)^{n(1+2+3+4+...+n)}\right]^\frac{1}{n^2}$$
$$=\lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)^{n(\frac {n^2+n}{2})}\right]^\frac{1}{n^2}$$
$$=\lim_{n \to \infty}(1+\frac 1 n)^{n(\frac1 2+\frac 1 {2n})}$$
$$=e^{\frac 1 2 + 0}=\sqrt e$$
As can be seen if $$a_{n} = \frac{(n + 1)(n + 2)\cdots (2n)}{n^{n}}$$ then the sequence in question is $a_{n}^{1/n}$. Now we can see that $$\frac{a_{n + 1}}{a_{n}} = \frac{(n + 2)(n + 3)\cdot (2n)(2n + 1)(2n + 2)}{(n + 1)^{n + 1}}\cdot\frac{n^{n}}{(n + 1)(n + 2) \cdots (2n)}$$ so that $$\frac{a_{n + 1}}{a_{n}} = \frac{2(2n + 1)}{n + 1}\cdot\left(\frac{n}{n + 1}\right)^{n} \to \frac{4}{e}$$ as $n \to \infty$ Hence the sequence $a_{n}^{1/n}$ also tends to $4/e$.
In general if $b_{n} = (a_{1}a_{2}\cdots a_{n})^{1/n}$ and $a_{n} \to L$ then $b_{n} \to L$.