Question

If ${{S}_{n}}=(1+{{3}^{-1}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}})$, then ${{S}_{\infty }}$ is equal to
A. $1$
B. $\dfrac{1}{2}$
C. $\dfrac{3}{2}$
D. None of these

Answer 1

Hint: In this question, we are to find the sum of the infinite series. Here we don’t know the type of the series. Then, we have another method to solve this series and to find the required sum. i.e., we can simplify the given expression, since the terms are related to one another. The simplification starts by multiplying the conjugate of the first term of the series on both sides. So, we can eliminate one of the terms from the series and it goes on till we get the required sum.

Formula used: Some of the important formulae are:
$\begin{align}
  & (a+b)(a-b)={{a}^{2}}-{{b}^{2}} \\
 & (a+b)(a+b)={{(a+b)}^{2}} \\
 & \Rightarrow {{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \\
 & (a-b)(a-b)={{(a-b)}^{2}} \\
 & \Rightarrow {{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} \\
\end{align}$

Complete step by step solution: Given series is
${{S}_{n}}=(1+{{3}^{-1}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}})$
Where we have the last term
On multiplying both sides of the series by the conjugate of the first term we have from the given series is $(1-{{3}^{-1}})$
I.e.,
$(1-{{3}^{-1}}){{S}_{n}}=(1-{{3}^{-1}})(1+{{3}^{-1}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}})$
Since we know that,
$(a-b)(a+b)={{a}^{2}}-{{b}^{2}}$ we can write $(1-{{3}^{-1}})(1+{{3}^{-1}})=(1-{{3}^{-2}})$.
On substituting, we get
$\begin{align}
  & (1-{{3}^{-1}}){{S}_{n}}=(1-{{3}^{-1}})(1+{{3}^{-1}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}}) \\
 & =(1-{{3}^{-2}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}}) \\
\end{align}$
Similarly, in the above equation, $(1-{{3}^{-2}})(1+{{3}^{-2}})=(1-{{3}^{-4}})$
So, on substituting, we get
$\begin{align}
  & (1-{{3}^{-1}}){{S}_{n}}=(1-{{3}^{-2}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}}) \\
 & (1-{{3}^{-1}}){{S}_{n}}=(1-{{3}^{-4}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}}) \\
\end{align}$
The above process goes on and we get
$\begin{align}
  & (1-{{3}^{-1}}){{S}_{n}}=(1-{{3}^{-{{2}^{n}}}})(1+{{3}^{-{{2}^{n}}}}) \\
 & \Rightarrow (1-{{3}^{-1}}){{S}_{n}}=1-{{({{3}^{-{{2}^{n}}}})}^{2}} \\
 & \Rightarrow (1-{{3}^{-1}}){{S}_{n}}=1-{{3}^{-{{2}^{n+1}}}} \\
 & \Rightarrow (1-\dfrac{1}{3}){{S}_{n}}=1-{{3}^{-{{2}^{n+1}}}} \\
\end{align}$
On simplifying, we get
$\begin{align}
  & (\dfrac{2}{3}){{S}_{n}}=1-{{3}^{-{{2}^{n+1}}}} \\
 & \Rightarrow {{S}_{n}}=\dfrac{3}{2}\left( 1-{{3}^{-{{2}^{n+1}}}} \right) \\
\end{align}$
Then, for the sum of the infinite terms, substitute $n=\infty $, then we get
$\begin{align}
  & {{S}_{n}}=\dfrac{3}{2}\left( 1-{{3}^{-{{2}^{n+1}}}} \right) \\
 & \Rightarrow {{S}_{\infty }}=\dfrac{3}{2}\left( 1-{{3}^{-{{2}^{\infty +1}}}} \right) \\
 & \Rightarrow {{S}_{\infty }}=\dfrac{3}{2}\left( 1-0 \right) \\
 & \therefore {{S}_{\infty }}=\dfrac{3}{2} \\
\end{align}$

Thus, Option (C) is correct.

Note: Here we need to remember that, the given series is unable to explain its type of progression. So, we used simplifying method which is possible for the given series. So, by using simple formulae, we get the required sum to the infinite series.