Correct Answer - Option 1 : Convergent
Concept:
Convergent or divergent series:
\(a_1\ +\ a_2\ +\ \dots a_n\ =\ \sum^{n}_{n=1}a_n\)
S1 = a1
S2 = a1 + a2
Similarly,
\(a_1\ +\ a_2\ +\ \dots a_n\ =\ \sum^{n}_{n=1}a_n\ =\ S_n\)
Where a1, a2, a3, ....an are the sequence of the real number.
Case:1 If \(\mathop {\lim }\limits_{n \to ∞ }S_n\ =\ S \ (finite)\) then \(\sum^{n}_{n=1}a_n\ =\ S_n\) is convergent.
Case: 2 If \(\mathop {\lim }\limits_{n \to ∞ }S_n\ =\ S \ (infinite)\) then \(\sum^{n}_{n=1}a_n\ =\ S_n\) is divergent.
Sum of n terms of GP:
The sum of n terms of a GP with first term ‘a’ and common ratio ‘r’ is given by:
\({S_n} = a\;\left( {\frac{{1 - {r^n}}}{{1 - r}}} \right)\) |r| < 1
Calculation:
Given series is
\(\frac{1}{3}\ +\ \frac{1}{3^2}\ +\ \frac{1}{3^3.}\ \dots\)
We can see that the above series is GP with a first term and common ratio of 1/3. Therefore, the sum of n term
\({S_n} = \frac{1}{3}\left[ {\frac{{1 - {(\frac{1}{3})^n}}}{{1 - (\frac{1}{3})}}} \right]\)
⇒ \({S_n} = \frac{1}{2}[1\ -\ \frac{1}{3^n}]\)
\(\mathop {\lim }\limits_{n \to ∞ }S_n\ =\ \mathop {\lim }\limits_{n \to ∞ } \frac{1}{2}[1\ -\ \frac{1}{3^n}]\)
⇒ \(\mathop {\lim }\limits_{n \to ∞ }S_n\ =\ \frac{1}{2}\) (∵ 1/∞ = 0)
Which is a finite value. Hence given series is convergent.