Answer
The numerical evidence suggests that the limit of the sequence:
$\mathop {\lim }\limits_{n \to + \infty } {\left( {\dfrac{{1 + n}}{{2n}}} \right)^n}$
is zero.
By the Squeezing Theorem for Sequences, we confirm that our conjecture is correct.
Work Step by Step
Using a calculating utility, we compute ${\left( {\dfrac{{1 + n}}{{2n}}} \right)^n}$ and list the values in the table below:
$\begin{array}{*{20}{c}}
n&{{{\left( {\dfrac{{1 + n}}{{2n}}} \right)}^n}}\\
1&{1.0}\\
2&{0.5625}\\
3&{0.2963}\\
4&{0.1525}\\
5&{0.07776}\\
6&{0.0394}\\
7&{0.01989}\\
8&{0.01002}\\
9&{0.005041}\\
{10}&{0.002533}\\
{11}&{0.001271}\\
{12}&{0.0006379}\\
{13}&{0.0003199}\\
{14}&{0.0001603}\\
{15}&{0.00008035}
\end{array}$
The results suggest that the limit of the sequence may be $0$. To confirm this, we evaluate the limit of ${a_n} = {\left( {\dfrac{{1 + n}}{{2n}}} \right)^n}$ as $n \to \infty $.
For $n \ge 3$, the following inequality is valid:
$0 \lt {\left( {\dfrac{{1 + n}}{{2n}}} \right)^n} \lt {\left( {\dfrac{{\dfrac{1}{2}n + n}}{{2n}}} \right)^n}$
So,
$0 \lt {\left( {\dfrac{{1 + n}}{{2n}}} \right)^n} \lt {\left( {\dfrac{{\dfrac{3}{2}n}}{{2n}}} \right)^n}$
$0 \lt {\left( {\dfrac{{1 + n}}{{2n}}} \right)^n} \lt {\left( {\dfrac{3}{4}} \right)^n}$
Taking the limit on all sides, we get
$0 \lt \mathop {\lim }\limits_{n \to + \infty } {\left( {\dfrac{{1 + n}}{{2n}}} \right)^n} \lt \mathop {\lim }\limits_{n \to + \infty } {\left( {\dfrac{3}{4}} \right)^n}$
$0 \lt \mathop {\lim }\limits_{n \to + \infty } {\left( {\dfrac{{1 + n}}{{2n}}} \right)^n} \lt 0$
By, the Squeezing Theorem for Sequences:
$\mathop {\lim }\limits_{n \to + \infty } {\left( {\dfrac{{1 + n}}{{2n}}} \right)^n} = 0$
This confirm our conjecture is correct.
