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Question: The sum of the series \(1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)...

The sum of the series 1+2(1+1n)+3(1+1n)2+...+∞1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty is given by
A) n2+1{n^2} + 1
B) n(n+1)n(n + 1)
C) n(1+1n)2n{\left( {1 + \dfrac{1}{n}} \right)^2}
D) n2{n^2}

Explanation

Solution

For an infinite series a1+a2+a3+...+∞{a_1} + {a_2} + {a_3} + ... + \infty , a quantity Sn=a1+a2+a3+...+an{S_n} = {a_1} + {a_2} + {a_3} + ... + {a_n}, which involves adding only the first n terms, is called a partial sum of the series. If Sn{S_n} approaches a fixed number S as n becomes larger and larger, the series is said to converge. In this case, S is called the sum of the series.
First, multiply infinite series with (1+1n)\left( {1 + \dfrac{1}{n}} \right). Then subtract both the series. First, simplify the left-hand side and then the right-hand side. Solve the equations with the help of LCM. Then Combine both sides and we will get the final answer. We will use the formula as stated below:
1+(1+1n)+(1+1n)2+...+∞=11−(1+1n)1 + \left( {1 + \dfrac{1}{n}} \right) + {\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty = \dfrac{1}{{1 - \left( {1 + \dfrac{1}{n}} \right)}}

Complete step-by-step solution:
The given infinite series is,
⇒Sn=1+2(1+1n)+3(1+1n)2+...+∞\Rightarrow {S_n} = 1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty
Let us multiply (1+1n)\left( {1 + \dfrac{1}{n}} \right) both sides.
Therefore,
⇒(1+1n)Sn=(1+1n)+2(1+1n)2+...+∞\Rightarrow \left( {1 + \dfrac{1}{n}} \right){S_n} = \left( {1 + \dfrac{1}{n}} \right) + 2{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty
Now, we will subtract Sn{S_n} and(1+1n)Sn\left( {1 + \dfrac{1}{n}} \right){S_n} .
Hence, the left-hand side is.
⇒Sn−(1+1n)Sn\Rightarrow {S_n} - \left( {1 + \dfrac{1}{n}} \right){S_n}
Solve the above equation.
⇒Sn−Sn−(1n)Sn\Rightarrow {S_n} - {S_n} - \left( {\dfrac{1}{n}} \right){S_n}
So, we get.
⇒−(1n)Sn\Rightarrow - \left( {\dfrac{1}{n}} \right){S_n}
Now, we will subtract 1+2(1+1n)+3(1+1n)2+...+∞1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty and(1+1n)+2(1+1n)2+...+∞\left( {1 + \dfrac{1}{n}} \right) + 2{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty .
Hence, the right-hand side is.
⇒[1+2(1+1n)+3(1+1n)2+...+∞]−[(1+1n)+2(1+1n)2+...+∞]\Rightarrow \left[ {1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{{\left( {1 + \dfrac{1}{n}} \right)}^2} + ... + \infty } \right] - \left[ {\left( {1 + \dfrac{1}{n}} \right) + 2{{\left( {1 + \dfrac{1}{n}} \right)}^2} + ... + \infty } \right]
Let us subtract the above step.
⇒1+2(1+1n)+3(1+1n)2+...+∞−(1+1n)−2(1+1n)2−...−∞\Rightarrow 1 + 2\left( {1 + \dfrac{1}{n}} \right) + 3{\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty - \left( {1 + \dfrac{1}{n}} \right) - 2{\left( {1 + \dfrac{1}{n}} \right)^2} - ... - \infty
Therefore, the answer will be.
⇒1+(1+1n)+(1+1n)2+...+∞\Rightarrow 1 + \left( {1 + \dfrac{1}{n}} \right) + {\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty
As we already know, the value of 1+(1+1n)+(1+1n)2+...+∞1 + \left( {1 + \dfrac{1}{n}} \right) + {\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty is 11−(1+1n)\dfrac{1}{{1 - \left( {1 + \dfrac{1}{n}} \right)}}
So, we put that value.
⇒11−(1+1n)\Rightarrow \dfrac{1}{{1 - \left( {1 + \dfrac{1}{n}} \right)}}
Let us simplify.
⇒11−(n+1n)\Rightarrow \dfrac{1}{{1 - \left( {\dfrac{{n + 1}}{n}} \right)}}
Let us take LCM.
⇒1(n−n−1n)\Rightarrow \dfrac{1}{{\left( {\dfrac{{n - n - 1}}{n}} \right)}}
So, we will get it.
⇒1(−1n)\Rightarrow \dfrac{1}{{\left( {\dfrac{{ - 1}}{n}} \right)}}
Therefore,
⇒−n\Rightarrow - n
Now, let us combine the left-hand side and right-hand side.
So, we will get it.
⇒−(1n)Sn=−n\Rightarrow - \left( {\dfrac{1}{n}} \right){S_n} = - n
Multiply both sides by−n - n .
⇒Sn=n2\Rightarrow {S_n} = {n^2}
Hence, the final answer is n2{n^2} .

Option D is the correct answer.

Note: The simplest way to express a result is as a series. To get the numerical value only we need to compute the value. For that, we have to remember the infinite series formula. Here, we use the formula1+(1+1n)+(1+1n)2+...+∞=11−(1+1n)1 + \left( {1 + \dfrac{1}{n}} \right) + {\left( {1 + \dfrac{1}{n}} \right)^2} + ... + \infty = \dfrac{1}{{1 - \left( {1 + \dfrac{1}{n}} \right)}}. Make sure that the LCM of two or more numbers is greater than or equal to the greatest number of given numbers.