For any two positive integers `x` and `y` `f(x,y)=1/((x+1)!)+1/((x+2)!)+1/((x+3)!)+……..+1/((x+y)!)`, then which of the following options is/are correct

Options

A. `f(x,y) le 1/x (1/(x!)-1/((x+y)!))`
B. `lim_(y to oo)(f(x,y)) le 1/(x!)`
C. `f(x,x) lt 1/((x-1)!)`
D. `f(2,2)` is equal to `2/3`

Explanation

To solve the problem, we need to evaluate the function \( f(x, y) \) defined as:

\[ f(x, y) = \frac{1}{(x+1)!} + \frac{1}{(x+2)!} + \frac{1}{(x+3)!} + \ldots + \frac{1}{(x+y)!} \]

### Step 1: Rewrite the Function
We can express the function in summation notation:

\[ f(x, y) = \sum_{i=1}^{y} \frac{1}{(x+i)!} \]

### Step 2: Analyze the Terms
Each term in the summation decreases rapidly as \( i \) increases because factorials grow very fast. Thus, we can analyze the behavior of the function as \( y \) increases.

### Step 3: Find an Upper Bound
To find an upper bound for \( f(x, y) \), we can observe that:

\[ \frac{1}{(x+i)!} < \frac{1}{(x+1)!} \quad \text{for all } i \geq 1 \]

This means that:

\[ f(x, y) < \sum_{i=1}^{y} \frac{1}{(x+1)!} = \frac{y}{(x+1)!} \]

### Step 4: Find a Lower Bound
For a lower bound, we can consider the last term in the summation:

\[ f(x, y) > \frac{1}{(x+y)!} \]

### Step 5: Establish Limits
As \( y \to \infty \), we can analyze the behavior of \( f(x, y) \):

\[ f(x, y) \approx \frac{1}{(x+1)!} + \frac{1}{(x+2)!} + \ldots \]

This series converges and can be approximated by the exponential series:

\[ \sum_{n=0}^{\infty} \frac{1}{n!} = e \]

### Step 6: Evaluate Specific Cases
To evaluate \( f(2, 2) \):

\[ f(2, 2) = \frac{1}{3!} + \frac{1}{4!} = \frac{1}{6} + \frac{1}{24} = \frac{4}{24} + \frac{1}{24} = \frac{5}{24} \]

### Conclusion
Based on the analysis, we can conclude that:

- \( f(x, y) \) is bounded above by \( \frac{y}{(x+1)!} \)
- \( f(x, y) \) is bounded below by \( \frac{1}{(x+y)!} \)