If `1/1^3 + (1+2)/(1^3+2^3)+(1+2+3)/(1^3+2^3+3^3) +.......n` terms then `lim_(n->oo) [S_n]`

Options

A. 2
B. 1
C. 3
D. 4

Answer

To solve the problem step-by-step, we will analyze the series given and find the limit as \( n \) approaches infinity.

### Step 1: Define the \( n \)-th term of the series
The \( n \)-th term \( T_n \) of the series can be expressed as:
\[ T_n = \frac{1 + 2 + 3 + \ldots + n}{1^3 + 2^3 + 3^3 + \ldots + n^3} \]

### Step 2: Use the formula for the sum of the first \( n \) natural numbers
The sum of the first \( n \) natural numbers is given by:
\[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \]

### Step 3: Use the formula for the sum of the cubes of the first \( n \) natural numbers
The sum of the cubes of the first \( n \) natural numbers is given by:
\[ 1^3 + 2^3 + 3^3 + \ldots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2 \]

### Step 4: Substitute these sums into \( T_n \)
Substituting the formulas from Steps 2 and 3 into \( T_n \):
\[ T_n = \frac{\frac{n(n + 1)}{2}}{\left(\frac{n(n + 1)}{2}\right)^2} \]

### Step 5: Simplify \( T_n \)
Simplifying \( T_n \):
\[ T_n = \frac{\frac{n(n + 1)}{2}}{\frac{n^2(n + 1)^2}{4}} = \frac{4}{n(n + 1)} \]

### Step 6: Find the sum \( S_n \)
The sum \( S_n \) is the summation of \( T_n \) from \( 1 \) to \( n \):
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} \frac{4}{k(k + 1)} \]

### Step 7: Simplify the summation using partial fractions
Using partial fractions, we can write:
\[ \frac{4}{k(k + 1)} = 4\left(\frac{1}{k} - \frac{1}{k + 1}\right) \]
Thus,
\[ S_n = 4\sum_{k=1}^{n} \left(\frac{1}{k} - \frac{1}{k + 1}\right) \]

### Step 8: Evaluate the telescoping series
The series telescopes:
\[ S_n = 4\left(1 - \frac{1}{n + 1}\right) = 4\left(\frac{n}{n + 1}\right) \]

### Step 9: Take the limit as \( n \) approaches infinity
Now we find the limit:
\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} 4\left(\frac{n}{n + 1}\right) = \lim_{n \to \infty} 4\left(\frac{1}{1 + \frac{1}{n}}\right) \]
As \( n \) approaches infinity, \( \frac{1}{n} \) approaches \( 0 \):
\[ \lim_{n \to \infty} S_n = 4 \cdot 1 = 4 \]

### Final Answer
Thus, the limit of the series as \( n \) approaches infinity is:
\[ \boxed{4} \]