Question

Evaluate the infinite series: \[ 1 + \frac{1}{3} + \frac{1.3}{3.6} + \frac{1.3.5}{3.6.9} + \dots \text{ to } \infty = \]

• A. \( \sqrt{5} \)
• B. \( \sqrt{6} \)
• C. \( \sqrt{15} \)
• D. \( \sqrt{3} \)

Hint:

For infinite series involving factorial and double factorial patterns, recognize standard summation results to quickly evaluate their limits.

Answer 1

The Correct Option is D

The given series is: \[ S = 1 + \frac{1}{3} + \frac{1.3}{3.6} + \frac{1.3.5}{3.6.9} + \dots \text{ to } \infty. \] Step 1: Identifying the pattern Observing the general term: \[ T_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{3 \cdot 6 \cdot 9 \cdots (3n)}. \] This is a standard series expansion for the function: \[ \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(3n)!!}. \] From known mathematical results, the sum of the given infinite series converges to: \[ \sqrt{3}. \] Step 2: Conclusion Thus, the given series evaluates to: \[ \boxed{\sqrt{3}}. \]

Answer 2

To evaluate the infinite series:

\[ S = 1 + \frac{1}{3} + \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \dots \text{ to } \infty \]

Step 1: Express the terms in factorial or double factorial form.
The numerator of the \(n\)-th term (starting from \(n=0\)) is the product of the first \(n\) odd numbers:
\[ 1 \cdot 3 \cdot 5 \cdots (2n-1) = (2n-1)!! \]
The denominator of the \(n\)-th term is:
\[ 3 \cdot 6 \cdot 9 \cdots 3n = 3^n (1 \cdot 2 \cdot 3 \cdots n) = 3^n n! \]

Step 2: Write the general term \(T_n\):
\[ T_n = \frac{(2n-1)!!}{3^n n!} \]

Step 3: Express double factorial \((2n-1)!!\) in terms of factorial:
\[ (2n-1)!! = \frac{(2n)!}{2^n n!} \]

Step 4: Substitute into \(T_n\):
\[ T_n = \frac{\frac{(2n)!}{2^n n!}}{3^n n!} = \frac{(2n)!}{(n!)^2 (2^n 3^n)} = \frac{(2n)!}{(n!)^2 (6^n)} \]

Step 5: Recognize the central binomial coefficient:
\[ \binom{2n}{n} = \frac{(2n)!}{(n!)^2} \]
So:
\[ T_n = \binom{2n}{n} \frac{1}{6^n} \]

Step 6: Write the series as:
\[ S = \sum_{n=0}^\infty \binom{2n}{n} \left(\frac{1}{6}\right)^n \]

Step 7: Use the generating function for central binomial coefficients:
\[ \sum_{n=0}^\infty \binom{2n}{n} x^n = \frac{1}{\sqrt{1-4x}}, \quad |x| < \frac{1}{4} \]

Step 8: Substitute \(x = \frac{1}{6}\) (which is less than \(\frac{1}{4}\)):
\[ S = \frac{1}{\sqrt{1 - \frac{4}{6}}} = \frac{1}{\sqrt{1 - \frac{2}{3}}} = \frac{1}{\sqrt{\frac{1}{3}}} = \sqrt{3} \]

Therefore, the value of the infinite series is:
\[ \boxed{\sqrt{3}} \]