Find the sum `Sigma_(r=1)^(oo) (r)/(r^4+1/4)`

Explanation

To find the sum \( S = \sum_{r=1}^{\infty} \frac{r}{r^4 + \frac{1}{4}} \), we will follow these steps:

### Step 1: Rewrite the term in the summation
We start with the term:
\[ T_r = \frac{r}{r^4 + \frac{1}{4}}. \]
We can rewrite the denominator:
\[ r^4 + \frac{1}{4} = r^4 + r^2 + \frac{1}{4} - r^2 = (r^2 + \frac{1}{2})^2 - r^2. \]
Thus, we have:
\[ T_r = \frac{r}{(r^2 + \frac{1}{2})^2 - r^2}. \]

### Step 2: Simplify the expression
We can express \( T_r \) in a different form:
\[ T_r = \frac{r}{\left(r^2 + \frac{1}{2}\right)^2 - r^2} = \frac{r}{\left(r^2 + \frac{1}{2} - r\right)\left(r^2 + \frac{1}{2} + r\right)}. \]

### Step 3: Partial fraction decomposition
We can use partial fractions to simplify \( T_r \):
\[ T_r = \frac{1}{2} \left( \frac{1}{r^2 - r + \frac{1}{2}} - \frac{1}{r^2 + r + \frac{1}{2}} \right). \]

### Step 4: Set up the summation
Now we can express the sum:
\[ S_n = \sum_{r=1}^{n} T_r = \frac{1}{2} \left( \sum_{r=1}^{n} \frac{1}{r^2 - r + \frac{1}{2}} - \sum_{r=1}^{n} \frac{1}{r^2 + r + \frac{1}{2}} \right). \]

### Step 5: Evaluate the limits
As \( n \) approaches infinity, we look at the behavior of the sums:
\[ \lim_{n \to \infty} S_n. \]
The terms will start to cancel out, and we will be left with the first and last terms of the series.

### Step 6: Find the final result
After evaluating the limit, we find:
\[ S = \lim_{n \to \infty} S_n = 1. \]

Thus, the sum \( S = \sum_{r=1}^{\infty} \frac{r}{r^4 + \frac{1}{4}} = 1 \).