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The series \(\frac{1}{3}\ +\ \frac{1}{3^2}\ +\ \frac{1}{3^3.}\ \dots\) is 
1. Convergent
2. Divergent
3. Oscillatory convergence
4. None of these

1 Answer

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Best answer
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. 

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