The limit of the series `underset(r=1)overset(n)sum (r)/(1+r^(2)+r^(4))` as n approaches infinity, is

Options

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

Answer

To solve the limit of the series

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

we will follow these steps:

### Step 1: Simplify the Denominator
The denominator can be rewritten as:

\[ 1 + r^2 + r^4 = r^4 + r^2 + 1 \]

This is a polynomial in \( r \).

### Step 2: Rewrite the Series
We can express the series as:

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

### Step 3: Factor the Denominator
Notice that \( r^4 + r^2 + 1 \) can be factored or analyzed for large \( r \):

For large \( r \), the term \( r^4 \) dominates. Therefore, we can approximate:

\[ 1 + r^2 + r^4 \approx r^4 \]

### Step 4: Rewrite the Fraction
Thus, we can rewrite the fraction as:

\[ \frac{r}{r^4 + r^2 + 1} \approx \frac{r}{r^4} = \frac{1}{r^3} \]

### Step 5: Rewrite the Series
Now, the series becomes:

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

### Step 6: Evaluate the Series
The series \( \sum_{r=1}^{n} \frac{1}{r^3} \) converges to \( \zeta(3) \) as \( n \to \infty \), but we are interested in the limit of the original series:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{r^3} \]

### Step 7: Calculate the Limit
As \( n \to \infty \), the series \( \sum_{r=1}^{n} \frac{1}{r^3} \) diverges, but we need to consider the average value of the series divided by \( n \):

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{r^3} \]

Using the integral test, we know that:

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

Thus,

\[ \lim_{n \to \infty} \frac{1}{n} \cdot \frac{n^2}{2} = \frac{1}{2} \]

### Final Answer
Therefore, the limit of the series as \( n \) approaches infinity is:

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