Sequences & Series Convergence Practice Questions, Quiz, and Step-by-Step Lesson - improve your math skills with focused questions and clear explanations.
Sequences & Series Convergence Practice Quiz with a Step-by-Step Interactive Lesson
Use the quiz at the top of the page to practice sequences and series convergence with the most important tools and patterns you will see in exams: sequence limits \(\lim_{n\to\infty} a_n\) (rational limits, exponential limits, and basic growth rates), the nth-term (divergence) test for series, geometric series and the key condition \(|r|<1\), alternating geometric series and quick sums, telescoping series using partial fractions, the p-series test (including the harmonic series), the comparison test and limit comparison test, the ratio test and root test (especially for factorials and exponentials), absolute vs. conditional convergence, and power series topics like radius of convergence and interval of convergence. If you want a refresher, click Start lesson to open a step-by-step guide with worked examples and quick checks.
How this sequences and series convergence practice works
1. Take the quiz: answer the sequences and series convergence questions at the top of the page.
2. Open the lesson (optional): review convergence tests, fast pattern recognition, and common sums with clear examples.
3. Retry: return to the quiz and apply the convergence rules immediately.
What you will learn in the sequences & series convergence lesson
Sequence limits & the divergence test
Limits of sequences: rational functions, polynomial degrees, and exponentials like \(\left(\tfrac{2}{3}\right)^n\)
Nth-term test: if \(\lim a_n ≠ 0\), then \(\sum a_n\) diverges
Common "trap" idea: \(\lim a_n=0\) is necessary but not sufficient for convergence
Geometric series & telescoping sums
Infinite geometric series: \(\sum ar^{n}\) converges when \(|r|<1\)
Fast sums: \(\sum_{n=0}^{\infty} r^n=\dfrac{1}{1-r}\) and \(\sum_{n=1}^{\infty} r^n=\dfrac{r}{1-r}\)
Telescoping series: rewrite terms to cancel and take a limit of partial sums
p-series, comparison tests, and growth
p-series test: \(\sum \dfrac{1}{n^p}\) converges if \(p>1\) and diverges if \(p\le 1\)
Comparison and limit comparison for matching difficult series to known benchmarks
Key intuition: exponentials beat polynomials, so terms like \(\dfrac{1}{n2^n}\) usually converge
Ratio/root tests & power series convergence
Ratio test and root test: ideal for factorials, exponentials, and power series
Absolute vs conditional convergence, especially for alternating series
Power series: find the radius of convergence \(R\) (and check endpoints for the interval)
Back to the quiz
When you are ready, return to the quiz at the top of the page and keep practicing sequences and series convergence.
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Series Convergence
Tests & power series
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Sequences & Series Convergence Lesson
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Lesson Overview
Lesson overview
Purpose: Build a clear understanding of sequences and series convergence so you can compute limits of sequences, apply the nth-term (divergence) test, recognize and sum geometric and telescoping series, and choose the right convergence test (p-series, comparison/limit comparison, ratio/root, and alternating series test). You’ll also learn how to find the radius of convergence (and check endpoints for the interval) for power series.
Success criteria
Compute limits of sequences such as rational and exponential sequences.
Use the nth-term test: if \(\lim_{n\to\infty} a_n≠ 0\), then \(\sum a_n\) diverges.
Recognize geometric series and use \(|r|<1\) to decide convergence.
Compute infinite geometric sums quickly.
Recognize a telescoping series and compute its sum using partial sums.
Classify \(\sum \dfrac{1}{n^p}\) using the p-series test and recognize the harmonic series.
Use comparison and limit comparison to match a hard series to a known benchmark.
Use the ratio test and root test (especially for factorials and power series).
Decide absolute vs conditional convergence for alternating series.
Find the radius of convergence \(R\) of a power series and check endpoints for the interval of convergence.
Key vocabulary
Sequence: an ordered list \((a_n)\). It converges if \(\lim a_n\) exists and is finite.
Series: a sum \(\sum a_n\). Its partial sums are \(S_n=\sum_{k=1}^{n} a_k\).
Converge / diverge: a series converges if \((S_n)\) approaches a finite limit; otherwise it diverges.
Nth-term (divergence) test: if \(\lim a_n≠ 0\) (or does not exist), the series \(\sum a_n\) diverges.
Geometric series: terms have constant ratio \(r\). Converges when \(|r|<1\).
p-series: \(\sum \dfrac{1}{n^p}\) converges if \(p>1\), diverges if \(p\le 1\).
Power series: \(\sum c_n (x-a)^n\). The radius \(R\) is found with ratio/root tests.
Quick pre-check
Pre-check 1: What is \(\lim_{n\to\infty}\dfrac{5n}{5n+1}\)?
Hint: Divide numerator and denominator by \(n\).
Pre-check 2: Does the series \(\sum_{n=1}^\infty \left(\tfrac12\right)^n\) converge or diverge?
Hint: It’s geometric with ratio \(r=\tfrac12\).
Sequence Limits
Sequence limits and the nth-term (divergence) test for series
Learning goal: Evaluate common limits of sequences and use the nth-term test to quickly spot series that must diverge.
Key idea
A sequence \((a_n)\) converges if \(\lim_{n\to\infty} a_n\) exists and is finite. Here are reliable patterns you should know:
Rational sequences: if degrees match, the limit is the ratio of leading coefficients. For example, \[ \lim_{n\to\infty}\frac{an+b}{cn+d}=\frac{a}{c}\quad (c≠ 0). \]
Polynomial growth: if the numerator degree is smaller than the denominator degree, the limit is \(0\). If it’s larger, the sequence typically grows without bound.
Exponential sequences: for \(r^n\):
If \(|r|<1\), then \(r^n\to 0\).
If \(r>1\), then \(r^n\to \infty\).
Example: \(\left(\tfrac{2}{3}\right)^n\to 0\), but \(\left(\tfrac{7}{4}\right)^n\to \infty\).
For series convergence, the first quick filter is the nth-term (divergence) test: \[ \text{If }\lim_{n\to\infty} a_n ≠ 0\text{ (or does not exist), then }\sum_{n=1}^{\infty} a_n \text{ diverges.} \] Important: \(\lim a_n = 0\) is necessary for convergence, but it does not guarantee convergence.
Worked example
Example: Evaluate \(\lim_{n\to\infty}\dfrac{3n+5}{3n+2}\). What does this imply about \(\sum_{n=1}^{\infty}\dfrac{3n+5}{3n+2}\)?
Divide by \(n\): \[ \frac{3n+5}{3n+2}=\frac{3+\frac{5}{n}}{3+\frac{2}{n}}\xrightarrow[n\to\infty]{}\frac{3}{3}=1. \] Since the terms do not go to \(0\), the series \(\sum_{n=1}^{\infty}\dfrac{3n+5}{3n+2}\) diverges by the nth-term test.
Try it
Try it 1: What is \(\lim_{n\to\infty}\dfrac{3n^2+1}{3n^2+4n}\)?
Hint: Divide numerator and denominator by \(n^2\).
Try it 2: Does the series \(\sum_{n=1}^{\infty}\dfrac{5n}{5n+1}\) converge or diverge?
Hint: \(\frac{5n}{5n+1}\to 1\). A series can’t converge if terms don’t go to \(0\).
Summary
Compute sequence limits using leading terms (especially for rational expressions in \(n\)).
For series, always check \(\lim a_n\) first: if it’s not \(0\), the series diverges.
Geometric & Telescoping
Geometric series sums and telescoping series
Learning goal: Recognize geometric series, compute their sums, and handle telescoping series with partial fractions.
Key idea
A geometric series has the form \(\sum ar^n\) (or a shifted version). It converges exactly when \(|r|<1\). The must-know sums:
\(\displaystyle \sum_{n=0}^{\infty} r^n=\frac{1}{1-r}\) for \(|r|<1\).
\(\displaystyle \sum_{n=1}^{\infty} r^n=\frac{r}{1-r}\) for \(|r|<1\).
A telescoping series is one where many terms cancel after rewriting: \[ a_n = b_n - b_{n+1}\quad \Rightarrow\quad \sum_{n=1}^{N} a_n = b_1 - b_{N+1}. \] This often happens after partial fraction decomposition.
Worked example
Example: What is \(\sum_{n=1}^{\infty} \dfrac{1}{2^n}\)?
This is geometric with \(r=\tfrac12\) and first term \(\tfrac12\): \[ \sum_{n=1}^{\infty}\left(\frac12\right)^n=\frac{\frac12}{1-\frac12}=1. \]
Try it
Try it 1: What is \(\sum_{n=0}^{\infty}\left(\tfrac13\right)^n\)?
Hint: Use \(\sum_{n=0}^{\infty} r^n=\dfrac{1}{1-r}\) with \(r=\tfrac13\).
Try it 2: What is \(\sum_{n=1}^{\infty}\dfrac{1}{n(n+3)}\)?
Geometric series converge when \(|r|<1\) and then you can sum them exactly.
Telescoping series become easy after rewriting terms so they cancel in the partial sum.
p-Series
p-series, the harmonic series, and quick classification
Learning goal: Recognize \(\sum \frac{1}{n^p}\) instantly and avoid the common p-series mistakes.
Key idea
The p-series test is one of the fastest convergence checks:
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^p}\) converges if \(p>1\).
\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^p}\) diverges if \(p\le 1\).
The harmonic series \(\sum \frac{1}{n}\) is the case \(p=1\), so it diverges.
You’ll also use p-series as benchmarks in the comparison test. If your terms behave like \(\frac{1}{n^p}\) for large \(n\), then the p-series test often tells you what happens.
Worked example
Example: Does the series \(\sum_{n=1}^{\infty}\frac{1}{n^{0.9}}\) converge or diverge?
This is a p-series with \(p=0.9\). Since \(p\le 1\), the series diverges.
Try it
Try it 1: Does the series \(\sum_{n=1}^{\infty}\frac{1}{n^{1/3}}\) converge or diverge?
Hint: \(\sum \frac{1}{n^p}\) converges only when \(p>1\).
Try it 2: Does the series \(\sum_{n=1}^{\infty}\frac{1}{n^{2}}\) converge or diverge?
Hint: For p-series, \(p>1\Rightarrow\) converges.
Summary
\(\sum \frac{1}{n^p}\) converges if \(p>1\), diverges if \(p\le 1\).
The harmonic series \(\sum \frac{1}{n}\) diverges (it’s \(p=1\)).
Comparison & Fast Sums
Comparison tests and “exponential beats polynomial”
Learning goal: Use comparison and limit comparison effectively, and compute common convergent sums like \(\sum n r^n\).
Key idea
The comparison test is a fast strategy when terms are positive:
If \(0\le a_n \le b_n\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
If \(0\le b_n \le a_n\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
A powerful intuition: exponentials beat polynomials. That’s why terms like \(\dfrac{1}{n2^n}\) often converge: the \(2^n\) in the denominator shrinks the terms very fast.
Also, this identity is extremely useful for sums: \[ \sum_{n=1}^{\infty} n r^n=\frac{r}{(1-r)^2}\quad (|r|<1). \] (You can derive it by differentiating the geometric series.)
Worked example
Example: Does \(\sum_{n=1}^{\infty}\dfrac{1}{n2^n}\) converge or diverge?
For \(n\ge 1\), we have \(n\ge 1\), so \(\dfrac{1}{n2^n}\le \dfrac{1}{2^n}\). The series \(\sum \dfrac{1}{2^n}\) is geometric with ratio \(\tfrac12\), so it converges. Therefore, \(\sum \dfrac{1}{n2^n}\) converges by comparison.
Try it
Try it 1: Does the series \(\sum_{n=1}^\infty \dfrac{1}{n2^n}\) converge or diverge?
Hint: Compare to \(\sum \frac{1}{2^n}\).
Try it 2: What is \(\sum_{n=1}^{\infty}\dfrac{n}{5^n}\)?
Hint: Use \(\sum_{n=1}^{\infty} n r^n=\dfrac{r}{(1-r)^2}\) with \(r=\tfrac15\).
Summary
Comparison works best when you can sandwich your series between simple “benchmark” series.
Know \(\sum n r^n=\dfrac{r}{(1-r)^2}\) for \(|r|<1\) to compute common convergent sums quickly.
Ratio & Root Tests
Ratio test, root test, and factorial/exponential series
Learning goal: Use the ratio test and root test confidently, especially for factorials, exponentials, and power-series-style terms.
Key idea
The ratio test looks at \[ L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|. \] Then:
If \(L<1\), the series \(\sum a_n\) converges absolutely.
If \(L>1\) (or \(L=\infty\)), the series diverges.
If \(L=1\), the test is inconclusive.
The root test uses \[ L=\limsup_{n\to\infty}\sqrt[n]{|a_n|}. \] It has the same conclusion rules as the ratio test and is especially nice for terms like \((\text{something})^n\).
Worked example
Example: Does the series \(\sum_{n=0}^{\infty}\dfrac{1}{n!}\) converge or diverge?
Let \(a_n=\dfrac{1}{n!}\). Then \[ \left|\frac{a_{n+1}}{a_n}\right| = \frac{1/(n+1)!}{1/n!} = \frac{1}{n+1} \xrightarrow[n\to\infty]{}0. \] So \(L=0<1\), and the series converges absolutely (in fact, it sums to \(e\)).
Try it
Try it 1: What is \(\lim_{n\to\infty}\dfrac{a_{n+1}}{a_n}\) for \(a_n=\left(\tfrac14\right)^n\)?
Hint: \(\frac{(1/4)^{n+1}}{(1/4)^n}=\tfrac14\).
Try it 2: What is \(\lim_{n\to\infty}\left(\tfrac{7}{4}\right)^n\)?
Hint: If \(r>1\), then \(r^n\to\infty\).
Summary
Ratio/root tests are perfect for factorials and exponential/power series terms.
For geometric sequences, \(\frac{a_{n+1}}{a_n}\) is the constant ratio \(r\).
Alternating Series
Alternating series test and absolute vs conditional convergence
Learning goal: Decide convergence for alternating series and identify when a convergent series is absolute or conditional.
Key idea
An alternating series often looks like \[ \sum_{n=1}^{\infty} (-1)^{n-1} b_n \quad \text{or} \quad \sum_{n=1}^{\infty} (-1)^{n} b_n, \] where \(b_n \ge 0\). The alternating series test says the series converges if:
\(b_n\) is eventually decreasing, and
\(\lim_{n\to\infty} b_n=0\).
A series converges absolutely if \(\sum |a_n|\) converges. If \(\sum a_n\) converges but \(\sum |a_n|\) diverges, then \(\sum a_n\) converges conditionally.
Worked example
Example: Does \(\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{1}{n^{0.9}}\) converge absolutely, converge conditionally, or diverge?
Let \(b_n=\dfrac{1}{n^{0.9}}\). Then \(b_n\) decreases and \(b_n\to 0\). So \(\sum (-1)^{n+1} b_n\) converges by the alternating series test. But \(\sum |(-1)^{n+1}b_n|=\sum \dfrac{1}{n^{0.9}}\) is a p-series with \(p=0.9\le 1\), so it diverges. Therefore the alternating series converges conditionally.
Try it
Try it 1: What is \(\sum_{n=1}^{\infty}(-1)^{n+1}\left(\tfrac12\right)^n\)?
Hint: It’s geometric with first term \(\tfrac12\) and ratio \(r=-\tfrac12\). Use \(S=\dfrac{a_1}{1-r}\).
Try it 2: The series \(\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{1}{n^{0.9}}\) is best described as:
Hint: Alternating test gives convergence, but \(\sum \frac{1}{n^{0.9}}\) diverges.
Absolute convergence means \(\sum |a_n|\) converges; conditional means \(\sum a_n\) converges but \(\sum |a_n|\) diverges.
Power Series
Power series convergence: radius and interval of convergence
Learning goal: Find the radius of convergence \(R\) using ratio/root tests and remember to check endpoints for the interval of convergence.
Key idea
A power series centered at \(a\) looks like \[ \sum_{n=0}^{\infty} c_n (x-a)^n. \] Usually, there is a radius \(R\) such that:
The series converges for \(|x-a|<R\).
The series diverges for \(|x-a|>R\).
At \(|x-a|=R\), you must test endpoints separately.
The ratio test often produces a condition like \(|x-a|<R\), and then \(R\) is the radius of convergence.
Worked example
Example: Find the radius of convergence of \(\sum_{n=1}^{\infty}\dfrac{x^n}{\sqrt{n}}\).
Use the root test (or ratio test): \[ \sqrt[n]{\left|\frac{x^n}{\sqrt{n}}\right|} = |x|\cdot \sqrt[n]{\frac{1}{\sqrt{n}}}. \] But \(\sqrt[n]{\frac{1}{\sqrt{n}}}=n^{-1/(2n)}\to 1\). So the limit is \(|x|\). The series converges when \(|x|<1\) and diverges when \(|x|>1\). Therefore, the radius of convergence is \(R=1\).
Try it
Try it 1: What is the radius of convergence of \(\sum_{n=1}^{\infty}\dfrac{x^n}{\sqrt n}\)?
Hint: Root/ratio tests lead to \(|x|<1\).
Try it 2: What is the radius of convergence of \(\sum_{n=0}^{\infty} x^n\)?
Hint: \(\sum x^n\) is geometric with ratio \(x\). It converges when \(|x|<1\).
Final recap
Nth-term test: if \(\lim a_n≠ 0\), then \(\sum a_n\) diverges.
Geometric series: converge when \(|r|<1\), sum with \(\frac{1}{1-r}\) (or \(\frac{r}{1-r}\) when starting at \(n=1\)).
Telescoping: rewrite to cancel in partial sums; take the limit.
p-series: \(\sum \frac{1}{n^p}\) converges if \(p>1\), diverges if \(p\le 1\).
Comparison/limit comparison: match to a known benchmark series.
Ratio/root tests: best for factorials, exponentials, and power series.
Alternating series: decreasing \(b_n\to 0\) implies convergence; decide absolute vs conditional by checking \(\sum |a_n|\).
Power series: find radius \(R\) and then test endpoints to get the interval of convergence.
Next step: Close this lesson and try your quiz again. If you miss a question, reopen the book and review the page that matches the convergence idea you need.
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