How do you find a formula for the sum n terms \[\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)\left( \dfrac{2}{n} \right)}\] and then find the limit as $n \to \infty $?

Hint: For simplifying the given sum \[\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)\left( \dfrac{2}{n} \right)}\], we can take $\dfrac{2}{n}$ outside the summation since it is a constant. On doing so, the summation will become \[\dfrac{2}{n}\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)}\]. Then using the property of the summation \[\sum{A+B}=\sum{A}+\sum{B}\], the summation \[\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)}\] will become \[\sum\limits_{i=1}^{n}{\left( 1 \right)}+\sum\limits_{i=1}^{n}{\left( \dfrac{i}{n} \right)}\]. For solving this, we have to take the common terms outside the second summation and use the formula for the summation of the first n natural numbers, given by $\dfrac{n\left( n+1 \right)}{2}$.

Complete step by step solution:
Let us consider the sum given in the above question as
$\Rightarrow {{S}_{n}}=\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)\left( \dfrac{2}{n} \right)}$
Since the variable of the limit in the above sum is $i$, $n$ is a constant for the given sum. This means that the factor of $\dfrac{2}{n}$ can be taken outside of the summation to get
$\Rightarrow {{S}_{n}}=\dfrac{2}{n}\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)}.........\left( i \right)$
Now, let us consider the new summation as
$\Rightarrow {{S}_{n}}'=\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)}........\left( ii \right)$
From the properties of summation, they know that \[\sum{A+B}=\sum{A}+\sum{B}\]. Therefore, the above summation can be written as
\[\Rightarrow {{S}_{n}}'=\sum\limits_{i=1}^{n}{\left( 1 \right)}+\sum\limits_{i=1}^{n}{\left( \dfrac{i}{n} \right)}\]
Taking $\dfrac{1}{n}$ outside the second summation in the above equation, we get
\[\Rightarrow {{S}_{n}}'=\sum\limits_{i=1}^{n}{\left( 1 \right)}+\dfrac{1}{n}\sum\limits_{i=1}^{n}{\left( i \right)}\]
Expanding both the above summation by putting the values of the variable $i$ from $1$ to $n$, we get
\[\Rightarrow {{S}_{n}}'=1+1+1+1+.....(\text{n times)}+\dfrac{1}{n}\left( 1+2+3+.....+n \right)\]
On adding $1$ $n$ times, we will get $n$. So the above summation becomes
\[\Rightarrow {{S}_{n}}'=n+\dfrac{1}{n}\left( 1+2+3+.....+n \right)\]
Now, in the above equation within the parenthesis, we have the sum of first n natural numbers. We know that the summation of first n natural numbers is equal to $\dfrac{n\left( n+1 \right)}{2}$. Therefore, the above equation now becomes
\[\begin{align}
  & \Rightarrow {{S}_{n}}'=n+\dfrac{1}{n}\dfrac{n\left( n+1 \right)}{2} \\
 & \Rightarrow {{S}_{n}}'=n+\dfrac{\left( n+1 \right)}{2} \\
 & \Rightarrow {{S}_{n}}'=\dfrac{2n+n+1}{2} \\
 & \Rightarrow {{S}_{n}}'=\dfrac{3n+1}{2} \\
\end{align}\]
Putting this in (ii) we get
$\begin{align}
  & \Rightarrow \dfrac{3n+1}{2}=\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)} \\
 & \Rightarrow \sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)}=\dfrac{3n+1}{2} \\
\end{align}$
Putting this in (i) we get
\[\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{2}{n}\left( \dfrac{3n+1}{2} \right) \\
 & \Rightarrow {{S}_{n}}=\dfrac{3n+1}{n} \\
\end{align}\]
Thus, the summation of the n terms of the given series reduces to \[\dfrac{3n+1}{n}\]. Now, according to the question, we consider the limit
$\begin{align}
  & \Rightarrow L=\displaystyle \lim_{n \to \infty }{{S}_{n}} \\
 & \Rightarrow L=\displaystyle \lim_{n \to \infty }\left( \dfrac{3n+1}{n} \right) \\
 & \Rightarrow L=\displaystyle \lim_{n \to \infty }\left( 3+\dfrac{1}{n} \right) \\
 & \Rightarrow L=3+\displaystyle \lim_{n \to \infty }\left( \dfrac{1}{n} \right) \\
\end{align}$
As the denominator of the above limit will increase infinitely, the value of the limit will become zero. So the required limit is obtained as
$\Rightarrow L=3$
Hence, the required summation is \[\dfrac{3n+1}{n}\] and the required limit is equal to $3$.

Note:
In the given sum \[\sum\limits_{i=1}^{n}{\left( 1+\dfrac{i}{n} \right)\left( \dfrac{2}{n} \right)}\], the variable of the summation is \[i\], and not \[n\]. It is a general misconception to consider \[n\] as the variable of the limit, but for the variable we must observe the summation of the symbol where the variable along with the upper and the lower limits are mentioned. For solving the limit as $n \to \infty $, we can also use the L-hospital’s rule since both the numerator and the denominator are tending to infinity.