Theory

Theory 1

Positive series

A series is called positive when its individual terms are positive, i.e. $a_n>0$ for all $n$.

The partial sum sequence $S_N$ is monotone increasing for a positive series.

By the monotonicity test for convergence of sequences, $S_N$ therefore converges whenever it is bounded above. If $S_N$ is not bounded above, then ${\sum }_{n=1}^{\infty }{a}_n$ diverges to $+\infty$.

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Another test, called the integral test, studies the terms of a series as if they represent rectangles with upper corner pinned to the graph of a continuous function.

To apply the test, we must convert the integer index variable $n$ in $a_n$ into a continuous variable $x$. This is easy when we have a formula for $a_n$ (provided it doesn’t contain factorials or other elements dependent on integrality).

Integral Test (IT)

Applicability: $f(x)$ must be:

• Continuous
• Positive
• Monotone decreasing

Test Statement:

$$\sum _{n=1}^{\infty }{a}_n\text{converges}⟺{\int }_1^{\infty }f(x)dx\text{converges}$$

Extra - Integral test: explanation

To show that integral convergence implies series convergence, consider the diagram:

This shows that ${\sum }_{n=2}^N{a}_n\le {\int }_1^{N}f(x)dx$ for any $N$. Therefore, if ${\int }_1^{\infty }f(x)dx$ converges, then ${\int }_1^{N}f(x)dx$ is bounded (independent of $N$) and so ${\sum }_{n=2}^N{a}_n$ is bounded by that inequality. But ${\sum }_{n=2}^N{a}_n=S_N-a_1$; so by adding $a_1$ to the bound, we see that $S_N$ itself is bounded, which implies that ${\sum }_{n=1}^{\infty }{a}_n$ converges.

To show that integral divergence implies series divergence, consider a similar diagram:

This shows that ${\sum }_{n=1}^{N-1}{a}_n\ge {\int }_1^{N}f(x)dx$ for any $N$. Therefore, if ${\int }_1^{\infty }f(x)dx$ diverges, then ${\int }_1^{N}f(x)dx$ goes to $+\infty$ as $N\to \infty$, and so ${\sum }_{n=1}^{N-1}{a}_n$ goes to $+\infty$ as well. So ${\sum }_{n=1}^{\infty }{a}_n$ diverges.

Notice: the picture shows $f(x)$ entirely above (or below) the rectangles. This depends upon $f(x)$ being monotone decreasing, as well as $f(x)>0$. (This explains the applicability conditions.)

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Next we use the integral test to evaluate the family of $p$-series, and later we can use $p$-series in comparison tests without repeating the work of the integral test.

$p$-series

A $p$-series is a series of this form: ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$

Convergence properties:

$$p>1:\text{series converges}p\le 1:\text{series diverges}$$

Extra - Proof of $p$-series convergence

(1) To verify the convergence properties of $p$-series, apply the integral test:

Applicability: verify it’s continuous, positive, decreasing.

Convert $n$ to $x$ to obtain the function $f(x)=\frac{1}{x^p}$.

Indeed $\frac{1}{x^p}$ is continuous and positive and decreasing as $x$ increases.

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(2) Apply the integral test.

Integrate, assuming $p\ne 1$:

$$\begin{array}{cc}{\int }_1^{\infty }\frac{1}{x^p}dx& \gg \gg \underset{R\to \infty }{\lim }{\frac{x^{p-1}}{p-1}|}_1^R\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }(\frac{R^{-p+1}}{-p+1}-\frac{1^{-p+1}}{-p+1})\end{array}$$

When $p>1$ we have ${\lim }_{R\to \infty }\frac{R^{-p+1}}{-p+1}=0$

When $p<1$ we have ${\lim }_{R\to \infty }\frac{R^{-p+1}}{-p+1}=\infty$

When $p=1$, integrate a second time:

$$\begin{array}{cc}{\int }_1^{\infty }\frac{1}{x}dx& \gg \gg \underset{R\to \infty }{\lim }\ln x{|}_1^R\\ & \\ & \gg \gg \underset{R\to \infty }{\lim }\ln R-\ln 1\gg \gg \infty \end{array}$$

Conclude: the integral converges when $p>1$ and diverges when $p\le 1$.

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We could instead immediately refer to the convergence results for $p$-integrals instead of reproving them here.

Theory 2

Direct Comparison Test (DCT)

Applicability: Both series are positive: $a_n>0$ and $b_n>0$.

Test Statement: Suppose $a_n\le b_n$ for large enough $n$. (Meaning: for $n\ge N$ with some given $N$.) Then:

• Smaller pushes up bigger:

$$\sum _{n=1}^{\infty }{a}_n\text{diverges}⟹\sum _{n=1}^{\infty }{b}_n\text{diverges}$$

• Bigger controls smaller:

$$\sum _{n=1}^{\infty }{b}_n\text{converges}⟹\sum _{n=1}^{\infty }{a}_n\text{converges}$$

Theory 3

Some series can be compared using the DCT after applying certain manipulations and tricks.

For example, consider the series ${\sum }_{n=2}^{\infty }\frac{1}{n^2-1}$. We suspect convergence because $a_n\approx \frac{1}{n^2}$ for large $n$. But unfortunately, $a_n>\frac{1}{n^2}$ always, so we cannot apply the DCT.

We could make some ad hoc arguments that do use the DCT, eventually:

Trick Method 1:

• Observe that for $n>1$ we have $\frac{1}{n^2-1}\le \frac{10}{n^2}$. (Check it!)
• But $\sum \frac{10}{n^2}$ converges, indeed its value is $10\cdot \sum \frac{1}{n^2}$, which is $\frac{10{\pi }^2}{6}$.
• So the series $\sum \frac{1}{n^2-1}$ converges.

Trick Method 2:

• Observe that we can change the letter $n$ to $n+1$ by starting the new $n$ at $n=1$.
• Then we have:

$$\sum _{n=2}^{\infty }\frac{1}{n^2-1}=\sum _{n=1}^{\infty }\frac{1}{{(n+1)}^2-1}=\sum _{n=1}^{\infty }\frac{1}{n^2+2n}$$

• This last series has terms smaller than $\frac{1}{n^2}$ so the DCT with $b_n=\frac{1}{n^2}$ (a convergent $p$-series) shows that the original series converges too.

These convoluted arguments suggest that a more general version of Comparison is possible.

Indeed, it is sufficient to compare the relative large-n behavior of the two series. We use the termwise ratios to estimate comparative behavior for increasing $n$.

Limit Comparison Test (LCT)

Applicability: Both series are positive: $a_n>0$ and $b_n>0$.

Test Statement: Suppose that ${\lim }_{n\to \infty }{\displaystyle \frac{a_n}{b_n}=L}$. Then: If $0<L<\infty$, i.e. finite non-zero, then:

$$\sum a_n\text{converges}⟺\sum b_n\text{converges}$$

Extra - LCT edge cases

If $L=0$ or $L=\infty$, we can still draw an inference, but only in one direction:

• If $L=0$:

$$\sum b_n\text{converges}⟹\sum a_n\text{converges}$$

• If $L=\infty$:

$$\sum b_n\text{diverges}⟹\sum a_n\text{diverges}$$

Extra - Limit Comparison Test explanation

Suppose $a_n/b_n\to L$ and $0<L<\infty$. Then for $n$ sufficiently large, we know $a_n/b_n<L+1$.

Doing some algebra, we get $a_n<(L+1)b_n$ for $n$ large.

If $\sum b_n$ converges, then $\sum (L+1)b_n$ also converges (constant multiple), and then the DCT implies that $\sum a_n$ converges.

Conversely: we also know that $b_n/a_n\to 1/L$, so $b_n<(1/L+1)a_n$ for all $n$ sufficiently large. Thus if $\sum a_n$ converges, $\sum (1/L+1)a_n$ also converges, and by the DCT again $\sum b_n$ converges too.

The cases with $L=0$ or $L=\infty$ are handled similarly.