The Main Idea 

An infinite series is what you get when you try to add infinitely many terms together. Since you can’t actually perform infinite addition, we define a series through the clever concept of partial sums—finite sums that we can calculate.

The core concept: For series [latex]\sum_{n=1}^{\infty} a_n[/latex], create the sequence of partial sums:

• [latex]S_1 = a_1[/latex]
• [latex]S_2 = a_1 + a_2[/latex]
• [latex]S_3 = a_1 + a_2 + a_3[/latex]
• And so on…

The series [latex]\sum_{n=1}^{\infty} a_n[/latex] converges to [latex]S[/latex] if and only if the sequence of partial sums [latex]{S_k}[/latex] converges to [latex]S[/latex].

To analyze any series, compute several partial sums and look for patterns. Does [latex]S_k[/latex] seem to approach a finite limit? Does it grow without bound? Does it oscillate?

The Main Idea 

The harmonic series [latex]\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots[/latex] is one of mathematics’ most deceptive examples—it looks like it should converge, but it actually diverges.

The proof strategy: Group terms cleverly to show the partial sums are unbounded:

• [latex]S_2 = 1 + \frac{1}{2}[/latex]
• [latex]S_4 > 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) = 1 + 2 \cdot \frac{1}{2}[/latex]
• [latex]S_8 > 1 + 3 \cdot \frac{1}{2}[/latex]
• In general: [latex]S_{2^j} > 1 + j \cdot \frac{1}{2}[/latex]

The key insight: Since [latex]1 + j \cdot \frac{1}{2} \to \infty[/latex] as [latex]j \to \infty[/latex], the partial sums are unbounded, so the series diverges.

The harmonic series demonstrates that having [latex]\lim_{n \to \infty} a_n = 0[/latex] is necessary but not sufficient for convergence. The terms must approach zero “fast enough”—the harmonic series shows the borderline between convergence and divergence.

The Main Idea 

Once you know that individual series converge, you can combine them using familiar algebraic operations. These rules work exactly like you’d expect from your experience with finite sums and limits.

The three fundamental rules:

• Sum Rule: [latex]\sum_{n=1}^{\infty} (a_n + b_n) = \sum_{n=1}^{\infty} a_n + \sum_{n=1}^{\infty} b_n[/latex]
• Difference Rule: [latex]\sum_{n=1}^{\infty} (a_n - b_n) = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{\infty} b_n[/latex]
• Constant Multiple Rule: [latex]\sum_{n=1}^{\infty} ca_n = c\sum_{n=1}^{\infty} a_n[/latex]

Important requirement: Both series must converge individually before you can apply these rules. These rules only apply to convergent series. If either series diverges, you cannot use these algebraic properties.

When facing a complex series, try to break it apart into simpler series that you can evaluate separately, then recombine using these rules.

The Main Idea 

Geometric series are the foundation of series theory because they have a clear pattern and a simple convergence test. Every geometric series has the form [latex]\sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + ar^3 + \cdots[/latex] where each term is the previous term multiplied by the common ratio [latex]r[/latex].

Geometric series give you an exact formula for the sum when they converge, making them the most useful and predictable type of series you’ll encounter.

The convergence rule is simple:

• If [latex]|r| < 1[/latex]: the series converges to [latex]\frac{a}{1-r}[/latex]
• If [latex]|r| \geq 1[/latex]: the series diverges

Why does this work? The partial sum formula [latex]S_k = \frac{a(1-r^k)}{1-r}[/latex] (for [latex]r \neq 1[/latex]) depends entirely on what happens to [latex]r^k[/latex] as [latex]k \to \infty[/latex]. When [latex]|r| < 1[/latex], we have [latex]r^k \to 0[/latex], so [latex]S_k \to \frac{a}{1-r}[/latex].

Look for a constant ratio between consecutive terms. The series doesn’t have to start with [latex]n=1[/latex] or have exponent [latex]n-1[/latex]—you can always rewrite it in standard form by factoring out appropriate terms.

Key technique: For series like [latex]\sum_{n=0}^{\infty} \left(\frac{2}{3}\right)^{n+2}[/latex], write out the first few terms to identify [latex]a[/latex] and [latex]r[/latex], then apply the convergence test.

The Main Idea 

Telescoping series are special because most terms cancel out when you compute partial sums, leaving only a few terms at the beginning and end. This makes them surprisingly easy to analyze despite looking complicated at first glance.

Telescoping series have the form [latex]\sum_{n=1}^{\infty}[b_n - b_{n+1}][/latex], where consecutive terms are designed to cancel each other out.

How the cancellation works: When you write out the partial sums:

• [latex]S_1 = b_1 - b_2[/latex]
• [latex]S_2 = (b_1 - b_2) + (b_2 - b_3) = b_1 - b_3[/latex]
• [latex]S_3 = (b_1 - b_2) + (b_2 - b_3) + (b_3 - b_4) = b_1 - b_4[/latex]

The middle terms vanish, leaving [latex]S_k = b_1 - b_{k+1}[/latex].

The series [latex]\sum_{n=1}^{\infty}[b_n - b_{n+1}][/latex] converges if and only if [latex]\lim_{k \to \infty} b_{k+1}[/latex] exists. If this limit is [latex]B[/latex], then the series converges to [latex]b_1 - B[/latex].

Look for series where you can use partial fractions or other algebraic techniques to rewrite terms as differences. For example, [latex]\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}[/latex].

Common telescoping forms:

• [latex]\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}[/latex]
• [latex]\frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right)[/latex]
• Functions like [latex]\cos(\frac{1}{n}) - \cos(\frac{1}{n+1})[/latex]

Telescoping series give you exact answers without requiring complex partial sum formulas. They’re one of the few types of series where you can find the exact sum rather than just determining convergence.