Prove that if converges, then converges.

Pf: Because for all n, the series converges by comparison with the convergent series . Furthermore, because we can write where both series on the right converge, which implies that converges.

Prove that if converges, then converges.

Pf: Because for all n, the series converges by comparison with the convergent series . Furthermore, because we can write where both series on the right converge, which implies that converges.

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If is positive, continuous, and decreasing for and , then and either both converge or both diverge.

To prove this theorem we first partition the interval into unit intervals. The total areas of the inscribed rectangles and the circumscribed triangles are as follows:

(inscribed area)

(circumscribed area)

The precise area under the graph of from to lies between the inscribed and circumscribed areas which implies . Using the nth partial sum, , we can write this inequality as . Now assuming that converges to L, it follows that for , . Consequently, is bounded and monotonic, and thereby by the Bounded Monotonic Sequence Theorem it converges. So, converges.

In this post I want to explain how a definite integral can be used to find the arc length of a curve because I think that it is fascinating that integrals have functions other than just finding the area under curves. Let us start with a definition:

A **RECTIFIABLE **curve is one that has a finite arc length. A sufficient condition for the graph of a function to be rectifiable between and is that be continuous on . Such a function is **continuously differentiable** on , and its graph on the interval is a **smooth curve**.

Now let’s consider a function that is continuously differentiable on the interval . We can approximate the graph of by *n *line segments whose endpoints are determined by the partition:

By letting and , we can approximate the length of the graph by the following:

This approximation becomes more and more precise as . Therefore the length of the graph is

Because exists for each in , the Mean Value Theorem guarantees the existence of in such that

Because is continuous on , it follows that is also continuous and therefore integrable on , which implies that =