# Convergence Proofs

Prove that if $\displaystyle\sum|a_n|$ converges, then $\displaystyle\sum a_n^2$ converges.

Pf: Because $0 \leq a_n^2 + |a_n| \leq |a_n^2 + a_n|$ for all n, the series $\displaystyle\sum_{n=1}^\infty (a_n^2 +|a_n|)$ converges by comparison with the convergent series $\displaystyle\sum_{n=1}^\infty |a_n^2 + a_n|$. Furthermore, because $a_n^2 = (a_n^2 + |a_n|)-|a_n|$ we can write $\displaystyle\sum_{n=1}^\infty a_n^2 = \displaystyle\sum_{n=1}^\infty (a_n^2 + |a_n|)-\displaystyle\sum_{n=1}^\infty |a_n|$ where both series on the right converge, which implies that $\sum a_n^2$ converges.

# The Integral Test

If $f$ is positive, continuous, and decreasing for $x\geq 1$and $a_{n} = f(n)$, then $\displaystyle\sum_{n=1}^\infty{a_n}$ and $\displaystyle\int_{1}^{\infty}f(x) dx$ either both converge or both diverge.

To prove this theorem we first partition the interval $[1, n]$ into $n-1$ unit intervals. The total areas of the inscribed rectangles and the circumscribed triangles are as follows:

$\displaystyle\sum_{i=2}^{n}f(i) = f(2) + f(3) + .... +f(n)$ (inscribed area)

$\displaystyle\sum_{i=1}^{n-1}f(i) = f(1) + f(2) + ... + f(n-1)$ (circumscribed area)

The precise area under the graph of $f$ from $x=1$ to $x=n$ lies between the inscribed and circumscribed areas which implies $\displaystyle\sum_{i=2}^{n}f(i) \leq \displaystyle\int_{1}^{n}f(x) dx \leq \displaystyle\sum_{i=1}^{n-1} f(i)$. Using the nth partial sum, $S_n = f(1) + f(2) +... + f(n)$, we can write this inequality as $S_n -f(1) \leq \displaystyle\int_1^n f(x) dx \leq S_{n-1}$. Now assuming that $\int_1^\infty f(x) dx$ converges to L, it follows that for $n \geq 1$, $S_{n} - f(1) \leq L\Rightarrow S_{n} \leq L + f(1)$ . Consequently, ${S_n}$ is bounded and monotonic, and thereby by the Bounded Monotonic Sequence Theorem it converges. So, $\sum a_n$ converges.

# Arc Length

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 $f$ to be rectifiable between $(a, f(a))$ and $(b, f(b))$ is that  $f'$ be continuous on $[a, b]$. Such a function is continuously differentiable on $[a, b]$, and its graph on the interval $[a, b]$ is a smooth curve.

Now let’s consider a function $y = f(x)$ that is continuously differentiable on the interval $[a, b]$. We can approximate the graph of $f$ by line segments whose endpoints are determined by the partition:

$a = x_{0} < x_{1} < x_{2} <....< x_{n} = b$

By letting $\Delta x_{i} = x_{i} - x_{i-1}$ and $\Delta y_{i} = y_{i} - y_{i-1}$, we can approximate the length of the graph by the following:

$\displaystyle S \approx \sum_{i=1}^{n} \sqrt{(x_{i} - x_{i-1})^2 +(y_{i} - y_{i-1})^2}$

$= \displaystyle \sum_{i=1}^{n} \sqrt{(\Delta x_{i})^2 + (\Delta y_{i})^2}$

$= \displaystyle \sum_{i=1}^{n} \sqrt{(\Delta x_{i})^2 +(\frac{\Delta y_{i}}{\Delta x_{i}})^2(\Delta x_{i})^2}$

$= \displaystyle \sum_{i=1}^{n} \sqrt{1 + (\frac{\Delta y_{i}}{\Delta x_{i}})^2}(\Delta x_{i})$

This approximation becomes more and more precise as $||\Delta|| \to 0 (n \to \infty)$. Therefore the length of the graph is $S = \displaystyle \lim_{||\Delta|| \to 0} \displaystyle \sum_{i=1}^{n} \sqrt{1 + (\frac{\Delta y_{i}}{\Delta x_{i}})^2}(\Delta x_{i})$

Because $f'(x)$ exists for each $x$ in $(x_{i-1}, x_i)$, the Mean Value Theorem guarantees the existence of $c_i$ in $(x_{i-1}, x_i)$ such that $f(x_i) - f(x_{i-1}) = f'(c_i)(x_i - x_{i-1})$

$\frac{\Delta y_i}{\Delta x_i} = f'(c_i)$

Because $f'$ is continuous on $[a, b]$, it follows that $\sqrt{1 + [f'(x)]^2}$ is also continuous and therefore integrable on $[a, b]$, which implies that $S = \displaystyle \lim_{||\Delta|| \to 0} \displaystyle \sum_{i=1}^{n} \sqrt{1 + [f'(c_i)]^2}(\Delta x_i)$                                                                               = $\int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx$