Problem 112 - Essential Calculus

Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The series \(\sum_{n=1}^{\infty} \frac{n}{1000(n+1)}\) diverges.

Step-by-Step Solution

Answer

The statement is false. The series \(\sum_{n=1}^{\infty} \frac{n}{1000(n+1)}\) converges.

Step 1: Write down the series

The series given is \(\sum_{n=1}^{\infty} \frac{n}{1000(n+1)}\)

Step 2: Simplify the series

The series can be simplified to \(\sum_{n=1}^{\infty} \frac{1}{1000} - \frac{1}{1000(n+1)}\)

Step 3: Apply the Divergence Test

We find the limit of \( \frac{1}{1000} - \frac{1}{1000(n+1)} \) as \(n\) approaches infinity, which is \( \lim_{n->\infty} \left( \frac{1}{1000} - \frac{1}{1000(n+1)} \right) = 0\). Since the limit is equal to zero, the Divergence Test gives us no information.

Step 4: Apply the test for Telescoping Series

Notice that the series can be written as a difference of fractions, which suggests that it can be a telescoping series. In a telescoping series the sum of the terms simplify significantly and in this case, it simplifies to \( \frac{1}{1000}-0 = \frac{1}{1000} \) which is finite, hence the series converges.