## Table of Contents

## Abel Summability

Consider the series \(1 - 2 + 3 - \dots.\) Convert it to a power series in \(x\) as follows: \(S(x) = 1 - 2x + 3x^2 - \dots\)

Sum this power series:

\[\begin{aligned} \int S(x) ~dx &= x - x^2 + x^3 - \dots = \frac{x}{1 + x} \newline S(x) &= \frac{d}{dx}\left[\frac{x}{1 + x}\right] = \frac1{(1 + x)^2} \end{aligned}\]

The Abel sum of the original series is defined as the \(\lim_{x \to 1-} S(x)\).

\[\text{Abel sum} = \lim_{x \to 1-}\left[\frac1{(1 + x)^2}\right] = \frac14\]

In general, oscillating series can be summed, but the order of the terms must not be changed. Because the Riemann rearrangement theorem says that if you get to reorder the terms you can make an oscillating series add up to anything.

## Summation of Series by Integration and Differentiation

\[\begin{aligned} S &= 1 + \frac14 + \frac19 + \frac1{16} + \dots\newline S(x) &= x + \frac{x^2}4 + \frac{x^3}9 + \dots \newline S’(x) &= 1 + \frac{x}2 + \frac{x^2}3 + \dots \newline &= \frac{-\ln(1 - x)}x \newline \newline S(x) &= -\int \frac{\ln(1 - x)}x ~dx \end{aligned}\]

One should know and recognize the series for the following expressions:

- \(\frac{1}{1 - x} = 1 + x + x^2 + x^3 + \dots \)
- \(-\ln(1 - x) = x + \frac{x^2}2 + \frac{x^3}3 + \dots \)
- \(\tan^{-1}x = x - \frac{x^3}3 + \frac{x^5}5 - \frac{x^7}7 + \dots \)
- \(e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \)
- \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \)
- \(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots \)
- \(\frac1{(1 + x)^2} = 1 - 2x + 3x^2 - \dots \)
- \(\frac1{\sqrt{1 + x}} = 1 - \frac12 x + \frac38 x^2 - \frac{15}{48}x^3 + \dots \) The coefficient of \(x^n\) is \[\frac{(2n - 1)!!}{(2n)!!} = \frac{1 \cdot 3 \cdot \dots \cdot (2n - 1)}{2 \cdot 4 \cdot \dots \cdot 2n}.\]

A useful relation: \( e^{i\theta} = \cos \theta + i \sin \theta \)

## Summation of Series by Adding Terms

Consider the series \(1 + \frac14 + \frac19 + \dots.\) We can sum it as follows: \(1.000 + .250 + .111 + .0625 + \dots\)

By taking enough terms we can obtain any desired accuracy. If the series is slow to converge it is sometimes possible to replace the higher terms by an integral:

Examination of the above diagram shows that

\[\begin{aligned} \frac1{3^2} + \frac1{4^2} + \dots &\approx \int_{2 \frac12}^\infty \frac1{x^2} ~dx \newline &\approx \frac1{2.5} = .4. \end{aligned}\]

The sum of our series is then Approximately; the exact value is \(\pi^2/6 = 1.6449…\) \(1.000 + .250 + .400 \approx 1.650.\) Since the error will be principally in the first terms some idea of its magnitude may be gotten by looking at the next best approximation:

\[\begin{aligned} \frac1{4^2} + \frac1{5^2} + \dots &\approx \int_{3 \frac12}^\infty \frac1{x^2} ~dx \newline &\approx .286 \end{aligned}\]

The new sum is \(1.000 + .250 + .111 + .286 \approx 1.647.\) We see that we have made an error of only \(.003\). We can guess from this that the third figure is probably about 6.

## Magnitude of Quantities

The following quantities are arranged in the order of their rate of rise for large \(x\): I prefer to write \(1/n\) for \(\epsilon\), or at least I did when I taught calculus. \[1 < \ln x < x^\epsilon < x < x^n < e^x \]

To ascertain their behavior for small \(y\), make the substitution \(y = \frac1x\).

The following approximations hold for small \(x\).

\[\begin{aligned} \sin x &\sim x \newline \cos x &\sim 1 - \frac{x^2}2 \newline \tan^{-1}x &\sim x \sim \tan x \newline \sinh x &\sim x \newline e^x &\sim 1 + x \newline (1 + x)^n &\sim 1 + nx \newline (1 + x)^{-1/2} &\sim 1 - \frac12 x \end{aligned}\]

## De L'Hospital's Rule

If as \(x \to a\), \(f(x)\) and \(g(x) \to 0\), then \[ \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f’(x)}{g’(x)}. \]

Example: \[ \lim_{x \to 0} \frac{\tan x}{[(1 + x)^3 - 1]} = \lim_{x \to 0} \frac{\sec^2 x}{3(1 + x)^2} = \frac13. \]

## Problems

- Sum:

\[\begin{gathered} \frac1{2!} + \frac2{3!} + \frac3{4!} + \dots \newline \frac{1}{1 \cdot 2} + \frac1{2 \cdot 3} + \frac1{3 \cdot 4} + \dots \end{gathered}\]

- Find \[\begin{gathered} \lim_{x \to 0} \frac{\sin x}{e^{3x} - 1} \newline \lim_{x \to 0} \frac{\ln(\frac12 + \frac1{2\sqrt{1+x^2}})}{e^{x^2}-\cos x^2}\end{gathered}\]