The value of limn→∞∏r=1n1+r2n2^2rn2 is equal to

Correct answer: (D) 4/e. Let y=limn→∞∏r=1n1+r2n2^2rn2 ⇒lny=limn→∞∑r=1n(2rn^2ln(1+r^2n^2)=∫012xln(1+x^2))/(dxNow, let 1+x^2=t⇒2xdx=dtSo, lny=∫12lntdt⇒lny=tlnt-t12⇒lny=2ln2−1=ln4e) ⇒y=4/e

Q116 / 160 · TS EAMCET 2020 (09 Sep Shift 1) · Mathematics

The value of $\lim_{n \to \infty} \prod_{r = 1}^{n} {(1 + \frac{r^{2}}{n^{2}})}^{\frac{2 r}{n^{2}}}$ is equal to

A
$\log (\frac{4}{e})$
B
$\log (\frac{2}{e})$
C
$\frac{2}{e}$
Correct answer
D
$\frac{4}{e}$

Solution

Let

$y = \lim_{n \to \infty} \prod_{r = 1}^{n} {(1 + \frac{r^{2}}{n^{2}})}^{\frac{2 r}{n^{2}}}$

$\Rightarrow \ln y = \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{2 r}{n^{2}} \ln (1 + \frac{r^{2}}{n^{2}})$

$= \int_{0}^{1} 2 x \ln (1 + x^{2}) d x$

Now, let $1 + x^{2} = t \Rightarrow 2 x d x = d t$

So, $\ln y = \int_{1}^{2} \ln (t) d t$

$\Rightarrow \ln y = {[t \ln (t) - t]}_{1}^{2}$

$\Rightarrow \ln y = 2 \ln 2 - 1 = \ln (\frac{4}{e})$

$\Rightarrow y = \frac{4}{e}$

Hence, the correct answer is option: 'D'.

The value of limn→∞∏r=1n1+r2n22rn2 is equal to

Solution

Relevant Questions

The value of $\lim_{n \to \infty} \prod_{r = 1}^{n} {(1 + \frac{r^{2}}{n^{2}})}^{\frac{2 r}{n^{2}}}$ is equal to