The value of `lim_(xtooo) {(1)/(3)+(2)/(21)+(3)/(91)+...+(n)/(n^4+n^2+1)}`, is

Options

A. 1
B. `1//2`
C. `1//3`
D. none of these

Answer

To find the value of the limit

\[ \lim_{n \to \infty} \left( \frac{1}{3} + \frac{2}{21} + \frac{3}{91} + \ldots + \frac{n}{n^4 + n^2 + 1} \right), \]

we will analyze the general term of the series, which is given by

\[ \frac{n}{n^4 + n^2 + 1}. \]

### Step 1: Simplifying the General Term

We start by simplifying the general term:

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

### Step 2: Finding the Limit of the General Term

As \( n \to \infty \), the term \( \frac{1}{n^2} \) and \( \frac{1}{n^4} \) approach 0. Thus, we can simplify further:

\[ \lim_{n \to \infty} \frac{1}{n^3(1 + \frac{1}{n^2} + \frac{1}{n^4})} = \lim_{n \to \infty} \frac{1}{n^3(1 + 0 + 0)} = \lim_{n \to \infty} \frac{1}{n^3} = 0. \]

### Step 3: Summing the Series

Now we need to sum the series:

\[ \sum_{k=1}^{n} \frac{k}{k^4 + k^2 + 1}. \]

As \( n \to \infty \), we can approximate:

\[ \sum_{k=1}^{n} \frac{k}{k^4 + k^2 + 1} \approx \sum_{k=1}^{n} \frac{1}{k^3}. \]

### Step 4: Evaluating the Series

The series \( \sum_{k=1}^{n} \frac{1}{k^3} \) converges as \( n \to \infty \). In fact, it converges to a finite value, specifically:

\[ \sum_{k=1}^{\infty} \frac{1}{k^3} = \zeta(3), \]

where \( \zeta \) is the Riemann zeta function.

### Step 5: Conclusion

Thus, the limit of the original series can be evaluated as:

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{k^4 + k^2 + 1} = \frac{1}{2}. \]

Therefore, the final answer is:

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