We are asked to find the value of the infinite sum $\sum_{k=1}^{\infty} \frac{1000k^2}{1+k^2}$.

1. Problem Description

We are asked to find the value of the infinite sum $\sum_{k=1}^{\infty} \frac{1000k^2}{1+k^2}$.

2. Solution Steps

First, we rewrite the expression inside the summation:

$\frac{1000k^2}{1+k^2} = \frac{1000(1+k^2-1)}{1+k^2} = \frac{1000(1+k^2)}{1+k^2} - \frac{1000}{1+k^2} = 1000 - \frac{1000}{1+k^2}$.

Therefore, we have

$\sum_{k=1}^{\infty} \frac{1000k^2}{1+k^2} = \sum_{k=1}^{\infty} (1000 - \frac{1000}{1+k^2}) = \sum_{k=1}^{\infty} 1000 - \sum_{k=1}^{\infty} \frac{1000}{1+k^2}$.

Since $\sum_{k=1}^{\infty} 1000$ diverges to infinity, we need to check if $\frac{1000k^2}{1+k^2}$ converges to $0$ as $k \to \infty$.

We have $\lim_{k \to \infty} \frac{1000k^2}{1+k^2} = \lim_{k \to \infty} \frac{1000}{1/k^2+1} = 1000 \neq 0$.

Therefore, by the divergence test, the series diverges.

3. Final Answer

$\infty$