The `r`th term of a series is given by `t_r=r/(1+r^2+r^4)`, then `lim(n->oo)sum_(r=1)^n(t_r)`

Options

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

Answer

To solve the problem, we need to find the limit of the sum of the series given by the \( r \)th term \( t_r = \frac{r}{1 + r^2 + r^4} \) as \( n \) approaches infinity.

### Step-by-Step Solution:

1. **Write the General Term**:
The \( r \)th term of the series is given by:
\[ t_r = \frac{r}{1 + r^2 + r^4} \]

2. **Factor the Denominator**:
We can factor the denominator \( 1 + r^2 + r^4 \):
\[ 1 + r^2 + r^4 = (r^2 + r + 1)(r^2 - r + 1) \]
Thus, we can rewrite \( t_r \) as:
\[ t_r = \frac{r}{(r^2 - r + 1)(r^2 + r + 1)} \]

3. **Rewrite the Term**:
We can separate \( t_r \) into two fractions:
\[ t_r = \frac{1}{2} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) \]

4. **Set Up the Sum**:
We need to find:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} t_r \]
This can be expressed as:
\[ \lim_{n \to \infty} \sum_{r=1}^{n} \left( \frac{1}{2} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) \right) \]

5. **Simplify the Sum**:
The sum can be simplified:
\[ \sum_{r=1}^{n} t_r = \frac{1}{2} \left( \sum_{r=1}^{n} \frac{1}{r^2 - r + 1} - \sum_{r=1}^{n} \frac{1}{r^2 + r + 1} \right) \]

6. **Observe Cancellation**:
Notice that the terms in the sum will cancel out:
\[ \sum_{r=1}^{n} \left( \frac{1}{r^2 - r + 1} - \frac{1}{r^2 + r + 1} \right) \]
This is a telescoping series, where most terms will cancel.

7. **Evaluate the Limit**:
As \( n \to \infty \), the remaining terms will lead to:
\[ \lim_{n \to \infty} \left( \frac{1}{2} \left( 1 - \frac{1}{n^2 + n + 1} \right) \right) = \frac{1}{2} \left( 1 - 0 \right) = \frac{1}{2} \]

8. **Final Answer**:
Thus, the limit is:
\[ \frac{1}{2} \]