June 30, 2026 - Calculus 2 - Finite and Infinite Series and Convergence Tests (Concise)

The professor plans to finish the series material this week.
There is a take home test on Thursday; it will be posted online for those attending orientation.
The test is currently due on Monday, but if material is not finished, it may be given Monday and due next Thursday.
There is no quiz on Thursday.

The geometric series formula is applicable only when the index starts at $1$ and the exponent is $n-1$ .
To adjust the index, if you subtract a value from the index, you must add that same value to $n$ in the expression (e.g., if $n=3$ , subtracting $2$ to get $1$ requires adding $2$ to the variable $n$ ).
Exponents can be simplified by peeling off factors: $\frac{2}{3}^{n+1}$ can be written as $\frac{2}{3}^{n} \times \frac{2}{3}^{1}$ .
A ratio $r$ must be between $-1$ and $1$ for convergence.
In the example using $2/3^{2}$ , the result is $4$ 9ths.
Checking if $34$ is in between $1$ and positive $1$ is necessary for convergence; in the example, $1+34 = 74$ .

A telescoping series collapses because terms in the middle cancel out, leaving only a few beginning and end terms.
Convergence is determined by the limit of the partial sums $S_{k}$ .
Example: Partial sums of $\frac{1}{n} - \frac{1}{n+1}$ results in a sum of $1 - \frac{1}{k+1}$ .
As $k \rightarrow \infty$ , the term $\frac{1}{k+1}$ goes to $0$ , meaning the series converges based on this remaining value.

If the limit of the sequence $a_{n}$ is any number other than $0$ (including infinity or minus infinity), the series diverges.
If the limit is $0$ , the test is inconclusive (e.g., the harmonic series $1/n$ has a limit of $0$ but is divergent).
One can use li talls or lows (treating it as a rational function) to find limits for this test.
Example: Since the limit of $\cos(\frac{1}{n})$ as $n \rightarrow \infty$ is $\cos(0) = 1$ , the series diverges by the divergence test because it does not equal $0$ .
Example: The limit of $e^{\frac{1}{n}}$ as $n \rightarrow \infty$ is $e^{0}$ , which causes it to diverges.

To use this test, let $a_{n} = f(n)$ . The function $f$ must be continuous, positive, and decreasing.
The series $\sum a_{n}$ and the improper integral $\int_{N}^{\infty} f(x)\,dx$ either both converge or both diverge.
Note: The value an integral converges to is typically different from the value the series converges to.
Example: $\int_{1}^{\infty} \frac{1}{x}\,dx = \ln(t)$ as $t \rightarrow \infty$ , which is infinity, proving the harmonic series diverges.

A P series is defined as $\sum \frac{1}{n^{p}}$ .
It converges if p > 1.
It diverges if $p \le 1$ .
If $p$ is negative, the divergence test proves the series is divergent (limit is infinity).
Example: For $\frac{1}{n^{4/5}}$ , the $p$ is $4/5$ , which is not bigger than $1$ , so the series diverges.

Compares an unknown series to a known one, like a P series.
If $a_{n} \le b_{n}$ for all $n \ge N$ and $\sum b_{n}$ converges, then $\sum a_{n}$ also converges.
If $a_{n} \ge b_{n}$ and $\sum b_{n}$ diverges, then $\sum a_{n}$ also diverges.
This test requires sequences to be non-negative, bounded, and monotone (eventually increasing or decreasing).

Question: Is manipulation the preferred way rather than writing out terms?
Response: Yes, as you do more with series, it becomes faster because it avoids writing out terms and calculating many individual values.
Question: Are these values on the unit circle?
Response: No pies involved; values are kept in terms of cosine.
Question: What does the divergence mean exactly? Is it getting close to one but never meets?
Response: It means adding $13$ every time makes the sum too large to capital; it gets closer to $13$ , and adding that repeatedly leads to infinity.
Question: How do you know when it ever converges?
Response: The divergence test cannot tell you if it converges. If the limit is $0$ , the test says nothing, and you must use another method.