Question

$\frac{1}{2.4}+\frac{1.3}{2.4.6}+\frac{1.3.5}{2.4.6.8}+\frac{1.3.5.7}{2.4.6.8.10}+\dots\dots\dots\infty$ is equal to -

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. 1

Answer 1

Answer: (C)

$\frac{1}{2}$

Answer 2

The given series is $S = \frac{1}{2.4}+\frac{1.3}{2.4.6}+\frac{1.3.5}{2.4.6.8}+\frac{1.3.5.7}{2.4.6.8.10}+\dots\dots\dots\infty$. The general term of the series can be written as $T_n = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n+2)}$. Using double factorials, the numerator is $(2n-1)!!$ and the denominator is $(2n+2)!!$. We can express these using factorials: $(2n-1)!! = \frac{(2n)!}{(2n)!!} = \frac{(2n)!}{2^n n!}$ $(2n+2)!! = 2^{n+1} (n+1)!$ Substituting these into $T_n$: $T_n = \frac{\frac{(2n)!}{2^n n!}}{2^{n+1} (n+1)!} = \frac{(2n)!}{2^{2n+1} n! (n+1)!}$ We can rewrite $T_n$ using the central binomial coefficient $\binom{2n}{n} = \frac{(2n)!}{n! n!}$: $T_n = \frac{1}{2} \cdot \frac{(2n)!}{2^{2n} n! (n+1)!} = \frac{1}{2} \cdot \frac{1}{n+1} \cdot \frac{(2n)!}{n! n!} \cdot \frac{1}{4^n} = \frac{1}{2} \frac{1}{n+1} \binom{2n}{n} \left(\frac{1}{4}\right)^n$ The series sum is $S = \sum_{n=1}^{\infty} T_n = \sum_{n=1}^{\infty} \frac{1}{2} \frac{1}{n+1} \binom{2n}{n} \left(\frac{1}{4}\right)^n = \frac{1}{2} \sum_{n=1}^{\infty} \binom{2n}{n} \frac{1}{n+1} \left(\frac{1}{4}\right)^n$. We use the known binomial series expansion: $\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^n}{n+1} = \frac{1 - \sqrt{1-4x}}{2x}$. Let $x = \frac{1}{4}$. Then $\sum_{n=0}^{\infty} \binom{2n}{n} \frac{(1/4)^n}{n+1} = \frac{1 - \sqrt{1-4(1/4)}}{2(1/4)} = \frac{1 - \sqrt{0}}{1/2} = 2$. Expanding the sum: $2 = \binom{0}{0} \frac{(1/4)^0}{0+1} + \sum_{n=1}^{\infty} \binom{2n}{n} \frac{(1/4)^n}{n+1} = 1 + \sum_{n=1}^{\infty} \binom{2n}{n} \frac{1}{4^n (n+1)}$. Therefore, $\sum_{n=1}^{\infty} \binom{2n}{n} \frac{1}{4^n (n+1)} = 2 - 1 = 1$. Substituting this back into the expression for $S$: $S = \frac{1}{2} \cdot (1) = \frac{1}{2}$.