To evaluate the mathematical expression shown in the image, let's break it down systematically:

The expression is: $\sum_{k=1}^\infty \left( \int_{n}^{n+1} \frac{1}{x^2 + 1} \, dx \right)$

Step 1: Solve the integral

The integrand is: $\frac{1}{x^2 + 1}$ The integral of $\frac{1}{x^2 + 1}$ is $\arctan(x)$. Thus: $\int_{n}^{n+1} \frac{1}{x^2 + 1} \, dx = \arctan(n+1) - \arctan(n)$

Step 2: Substitute the result into the summation

The summation becomes: $\sum_{k=1}^\infty \left(\arctan(n+1) - \arctan(n)\right)$

Step 3: Observe the telescoping series

The terms in the summation cancel out as follows: $\arctan(2) - \arctan(1) + \arctan(3) - \arctan(2) + \arctan(4) - \arctan(3) + \cdots$

The cancellation results in: $\lim_{n \to \infty} \arctan(n+1) - \arctan(1)$

Step 4: Evaluate the limit

As $n \to \infty$, $\arctan(n+1) \to \frac{\pi}{2}$, since the arctangent of a very large number approaches $\frac{\pi}{2}$. Therefore: $\lim_{n \to \infty} \arctan(n+1) - \arctan(1) = \frac{\pi}{2} - \arctan(1)$

The value of $\arctan(1)$ is $\frac{\pi}{4}$. Thus: $\frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

Final Answer:

$\boxed{\frac{\pi}{4}}$

Would you like further explanation or have any related questions?

Related Questions:

1. How do telescoping series simplify summations?
2. What is the significance of the arctangent function in integral calculus?
3. Could you evaluate $\sum_{k=1}^\infty \arctan(k)$ without telescoping terms?
4. How can improper integrals relate to infinite series?
5. How does the limit of $\arctan(x)$ behave as $x$ approaches infinity?

Tip:

When working with infinite series, always check for telescoping patterns—they can simplify evaluation significantly.