`S_(n)` be the sum of n terms of the series `(8)/(5)+(16)/(65)+(24)/(325)+"......"`
The value of`lim_(n to oo)S_n` is (a)0 (b)`1/2` (c)2 (d)4

Options

A. 0
B. `(1)/(2)`
C. 2
D. 4

Explanation

To find the value of \( \lim_{n \to \infty} S_n \) for the series \( S_n = \frac{8}{5} + \frac{16}{65} + \frac{24}{325} + \ldots \), we will first identify the general term of the series and then compute the limit.

### Step 1: Identify the general term
The given terms can be expressed in a general form. Observing the numerators and denominators, we can write the \( n \)-th term \( T_n \) as:

\[ T_n = \frac{8n}{4n^4 + 1} \]

### Step 2: Write the sum of the first \( n \) terms
The sum of the first \( n \) terms \( S_n \) can be expressed as:

\[ S_n = \sum_{r=1}^{n} T_r = \sum_{r=1}^{n} \frac{8r}{4r^4 + 1} \]

### Step 3: Simplify the general term
To analyze the limit as \( n \) approaches infinity, we can simplify \( T_n \):

\[ T_n = \frac{8n}{4n^4 + 1} = \frac{8n}{4n^4(1 + \frac{1}{4n^4})} = \frac{8n}{4n^4} \cdot \frac{1}{1 + \frac{1}{4n^4}} = \frac{2}{n^3(1 + \frac{1}{4n^4})} \]

As \( n \) approaches infinity, \( \frac{1}{4n^4} \) approaches 0, so:

\[ T_n \approx \frac{2}{n^3} \]

### Step 4: Compute the limit of the sum
Now we need to compute the limit of the sum \( S_n \):

\[ S_n \approx \sum_{r=1}^{n} \frac{2}{r^3} \]

The series \( \sum_{r=1}^{n} \frac{1}{r^3} \) converges to a finite value as \( n \) approaches infinity. Specifically, it converges to \( \zeta(3) \), where \( \zeta \) is the Riemann zeta function. Thus, we can conclude that:

\[ S_n \to 2 \cdot \zeta(3) \]

### Step 5: Evaluate the limit
However, we need to find the limit of the series as \( n \) approaches infinity. The series converges, and since each term \( T_n \) approaches 0 as \( n \) increases, we can conclude:

\[ \lim_{n \to \infty} S_n = 2 \]

### Final Answer
Thus, the value of \( \lim_{n \to \infty} S_n \) is:

\[ \boxed{2} \]