Let `f(r) = sum_(j=2)^(2008) (1)/(j^(r)) = (1)/(2^(r))+(1)/(3^(r))+"…."+(1)/(2008^(r))`. Find `sum_(k=2)^(oo) f(k)`

Answer

To solve the problem, we need to find the sum \( \sum_{k=2}^{\infty} f(k) \) where \( f(r) = \sum_{j=2}^{2008} \frac{1}{j^r} \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(r) = \sum_{j=2}^{2008} \frac{1}{j^r} = \frac{1}{2^r} + \frac{1}{3^r} + \ldots + \frac{1}{2008^r} \]

2. **Write the sum we need to evaluate**:
\[ \sum_{k=2}^{\infty} f(k) = \sum_{k=2}^{\infty} \left( \frac{1}{2^k} + \frac{1}{3^k} + \ldots + \frac{1}{2008^k} \right) \]

3. **Rearranging the sum**:
We can interchange the order of summation:
\[ \sum_{k=2}^{\infty} f(k) = \sum_{j=2}^{2008} \sum_{k=2}^{\infty} \frac{1}{j^k} \]

4. **Evaluate the inner sum**:
The inner sum \( \sum_{k=2}^{\infty} \frac{1}{j^k} \) is a geometric series with first term \( \frac{1}{j^2} \) and common ratio \( \frac{1}{j} \):
\[ \sum_{k=2}^{\infty} \frac{1}{j^k} = \frac{\frac{1}{j^2}}{1 - \frac{1}{j}} = \frac{\frac{1}{j^2}}{\frac{j-1}{j}} = \frac{1}{j^2} \cdot \frac{j}{j-1} = \frac{1}{j(j-1)} \]

5. **Substituting back into the sum**:
Now we substitute this back into our sum:
\[ \sum_{k=2}^{\infty} f(k) = \sum_{j=2}^{2008} \frac{1}{j(j-1)} \]

6. **Simplifying the sum**:
The term \( \frac{1}{j(j-1)} \) can be rewritten using partial fractions:
\[ \frac{1}{j(j-1)} = \frac{1}{j-1} - \frac{1}{j} \]
Therefore, the sum becomes:
\[ \sum_{j=2}^{2008} \left( \frac{1}{j-1} - \frac{1}{j} \right) \]

7. **Recognizing the telescoping series**:
This is a telescoping series:
\[ \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \ldots + \left( \frac{1}{2007} - \frac{1}{2008} \right) \]
Most terms cancel out, leaving:
\[ 1 - \frac{1}{2008} \]

8. **Final Calculation**:
Thus, we have:
\[ \sum_{k=2}^{\infty} f(k) = 1 - \frac{1}{2008} = \frac{2008 - 1}{2008} = \frac{2007}{2008} \]

### Final Answer:
\[ \sum_{k=2}^{\infty} f(k) = \frac{2007}{2008} \]