Question

$S_n = \frac{1}{2.4} + \frac{1.3}{2.4.6} + \frac{1.3.5}{2.4.6.8} + ...$

Answer 1

1/2

Answer 2

The given series is $S_n = \frac{1}{2.4} + \frac{1.3}{2.4.6} + \frac{1.3.5}{2.4.6.8} + ...$

Let's write the general term of the series, $T_k$.
The numerator of the $k$-th term (for $k \ge 1$) is the product of the first $k$ odd numbers: $1 \cdot 3 \cdot 5 \cdot \dots \cdot (2k-1)$.
The denominator of the $k$-th term is the product of the first $k+1$ even numbers: $2 \cdot 4 \cdot 6 \cdot \dots \cdot (2k+2)$.
So, $T_k = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2k-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2k+2)}$.

We can rewrite the numerator and denominator using factorials.
Numerator: $1 \cdot 3 \cdot 5 \cdot \dots \cdot (2k-1) = \frac{(2k)!}{2 \cdot 4 \cdot \dots \cdot (2k)} = \frac{(2k)!}{2^k k!}$.
Denominator: $2 \cdot 4 \cdot 6 \cdot \dots \cdot (2k+2) = 2^{k+1} (1 \cdot 2 \cdot 3 \cdot \dots \cdot (k+1)) = 2^{k+1} (k+1)!$.

So, the $k$-th term $T_k$ is:
$T_k = \frac{\frac{(2k)!}{2^k k!}}{2^{k+1} (k+1)!} = \frac{(2k)!}{2^k k! \cdot 2^{k+1} (k+1)!} = \frac{(2k)!}{2^{2k+1} k! (k+1)!}$
We know that $\binom{2k}{k} = \frac{(2k)!}{k! k!}$.
So, $T_k = \frac{1}{2^{2k+1}} \frac{1}{k+1} \frac{(2k)!}{k! k!} = \frac{1}{2^{2k+1}} \frac{1}{k+1} \binom{2k}{k}$.
$T_k = \frac{1}{2 \cdot 4^k} \frac{1}{k+1} \binom{2k}{k}$.

The sum $S_n$ (which is an infinite series, let's call it $S$) is:
$S = \sum_{k=1}^{\infty} T_k = \sum_{k=1}^{\infty} \frac{1}{2 \cdot 4^k} \frac{1}{k+1} \binom{2k}{k}$.
$S = \frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k+1} \binom{2k}{k} \left(\frac{1}{4}\right)^k$.

This series is related to the generating function for Catalan numbers. The $k$-th Catalan number is $C_k = \frac{1}{k+1} \binom{2k}{k}$.
The generating function for Catalan numbers is given by:
$G(x) = \sum_{k=0}^{\infty} C_k x^k = \sum_{k=0}^{\infty} \frac{1}{k+1} \binom{2k}{k} x^k$.
It is a known result that $G(x) = \frac{1 - \sqrt{1-4x}}{2x}$.

We need to evaluate the sum $\sum_{k=1}^{\infty} \frac{1}{k+1} \binom{2k}{k} \left(\frac{1}{4}\right)^k$.
This sum is $G\left(\frac{1}{4}\right)$ excluding the $k=0$ term.
The $k=0$ term of $G(x)$ is $C_0 x^0 = \frac{1}{0+1} \binom{0}{0} (x^0) = 1 \cdot 1 \cdot 1 = 1$.
So, $\sum_{k=1}^{\infty} \frac{1}{k+1} \binom{2k}{k} \left(\frac{1}{4}\right)^k = G\left(\frac{1}{4}\right) - C_0$.

Let's substitute $x = \frac{1}{4}$ into $G(x)$:
$G\left(\frac{1}{4}\right) = \frac{1 - \sqrt{1-4\left(\frac{1}{4}\right)}}{2\left(\frac{1}{4}\right)} = \frac{1 - \sqrt{1-1}}{\frac{1}{2}} = \frac{1 - \sqrt{0}}{\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$.

Now, substitute this back into the expression for $S$:
$S = \frac{1}{2} \left( G\left(\frac{1}{4}\right) - C_0 \right) = \frac{1}{2} (2 - 1) = \frac{1}{2} (1) = \frac{1}{2}$.

The sum of the series is $\frac{1}{2}$.