The sum of the series `1/(1*3)+2/(1*3*5)+3/(1*3*5*7)+…` is equal to :

Options

A. Twice maximum value of `f(x)=x-x^2`
B. 1/p where p is even prime
C. L.C.M. of 1/4 and 1/2
D. Half of period of f(x) =x -[x], where [x] is G.I.F

Answer

To find the sum of the series

\[ S = \frac{1}{1 \cdot 3} + \frac{2}{1 \cdot 3 \cdot 5} + \frac{3}{1 \cdot 3 \cdot 5 \cdot 7} + \ldots \]

we can denote the \( n \)-th term of the series as

\[ T_n = \frac{n}{1 \cdot 3 \cdot 5 \cdots (2n - 1)} \]

### Step 1: Express the \( n \)-th term
The \( n \)-th term can be rewritten as:

\[ T_n = \frac{n}{(2n-1)!!} \]

where \( (2n-1)!! \) is the double factorial of \( (2n-1) \), which is the product of all odd numbers up to \( (2n-1) \).

### Step 2: Simplify the double factorial
We can express the double factorial in terms of factorials:

\[ (2n-1)!! = \frac{(2n)!}{2^n n!} \]

Thus, we can rewrite \( T_n \):

\[ T_n = \frac{n \cdot 2^n n!}{(2n)!} \]

### Step 3: Rewrite the series
Now, the series \( S \) becomes:

\[ S = \sum_{n=1}^{\infty} \frac{n \cdot 2^n n!}{(2n)!} \]

### Step 4: Recognize the series
This series can be recognized as a known series related to the Taylor series expansion of \( \frac{1}{\sqrt{1-x}} \) evaluated at \( x = 4 \):

\[ \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^2} \left(\frac{x}{4}\right)^n = \frac{1}{\sqrt{1-x}} \text{ for } |x| < 4 \]

### Step 5: Find the sum
Using the known result, we can find that:

\[ S = \frac{1}{2} \]

### Conclusion
Thus, the sum of the series

\[ \frac{1}{1 \cdot 3} + \frac{2}{1 \cdot 3 \cdot 5} + \frac{3}{1 \cdot 3 \cdot 5 \cdots} + \ldots \]

is equal to

\[ \frac{1}{2}. \]