CEGEP Calculus II — Sequences and Series Worksheet

All 11 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Sequences and Their Limits

   Decide whether each sequence converges. Where it does, find the limit; where it does not, say what the terms do instead.

   1. $a_n={\displaystyle \frac{3n^2-n}{5n^2+4}}$
   2. $b_n={\displaystyle \frac{\ln n}{\sqrt{n}}}$
   3. $c_n={(1+{\displaystyle \frac{2}{n}})}^n$
   4. $d_n={\displaystyle \frac{(-1{)}^{n}n}{n+1}}$
2. Q2Series, Partial Sums and the General Term

   A desalination rig rinses a filter repeatedly. After $n$ rinse cycles the total mass of salt it has removed is $$S_n=\frac{3n}{n+2}\text{grams},n\ge 1.$$

   1. Find the mass $a_n$ removed by the $n$th cycle alone, as a formula valid for every $n\ge 1$.
   2. Find ${\displaystyle {\sum }_{n=1}^{\infty }{a}_n}$.
   3. Interpret the answer to (b) for the operator of the rig.
3. Q3Series, Partial Sums and the General Term

   For each series, write a formula for the general term $a_n$ valid for $n\ge 1$, then state whether the Divergence Test (the $n$th term test) settles the series — and if it does, give the conclusion.

   1. ${\displaystyle \frac{2}{5}}-{\displaystyle \frac{4}{7}}+{\displaystyle \frac{6}{9}}-{\displaystyle \frac{8}{11}}+\cdots$
   2. ${\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{12}}+{\displaystyle \frac{1}{20}}+\cdots$
   3. ${\displaystyle \frac{3}{1}}+{\displaystyle \frac{9}{4}}+{\displaystyle \frac{27}{9}}+{\displaystyle \frac{81}{16}}+\cdots$
4. Q4Geometric and Telescoping Series

   Find the exact sum of each convergent series, naming the type you are using.

   1. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{2^{n+1}}{3^{n-1}}}$
   2. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{2}{4n^2-1}}$
5. Q5The Integral Test

   Apply the Integral Test to each series. State and check the hypotheses on the companion function before integrating, and evaluate the improper integral.

   1. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{n}{(n^2+3{)}^2}}$
   2. ${\displaystyle {\sum }_{n=2}^{\infty }\frac{1}{n\ln n}}$
6. Q6The Comparison and Limit Comparison Tests

   Settle each series. Name which of the two tests you use, state the comparison series and why it converges or diverges, and show the inequality or the limit that justifies the conclusion.

   1. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{5}{2^n+n}}$
   2. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{\sqrt{n}+1}{n^2-n+3}}$
   3. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{n^2+2}{n^3+5n}}$
7. Q7The Ratio and Root Tests

   Apply the Ratio Test or the Root Test — whichever the shape of the general term invites — and state the limit $L$ and the conclusion.

   1. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{4^n}{n!}}$
   2. ${\displaystyle {\sum }_{n=1}^{\infty }{\left(\frac{3n+2}{2n-1}\right)}^n}$
   3. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{(2n)!}{(n!{)}^2{5}^n}}$
8. Q8Alternating Series
   1. Show that ${\displaystyle {\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}n}{n^2+4}}$ converges by the Alternating Series Test. Check both hypotheses, and comment on what happens between $n=1$ and $n=2$.
   2. The series ${\displaystyle {\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n^3}}$ converges. How many terms of it guarantee an approximation of the sum with error less than $0.001$? Justify with the alternating series remainder estimate.
9. Q9Absolute and Conditional Convergence

   Classify each series as absolutely convergent, conditionally convergent, or divergent. Justify each classification by naming the test applied to $\sum |a_n|$ and, where it is needed, the test applied to $\sum a_n$ itself.

   1. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{\sqrt{n+1}}}$
   2. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{n^2+1}}$
   3. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{n\cos (n\pi )}{n+6}}$
   4. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{(-2{)}^n}{n!}}$
10. Q10Selecting a Convergence Test

    For each series below: name the one test you would apply and say in a sentence why the shape of the general term invites it; name the stated alternative and say why it fails or is inconclusive; then give the conclusion. Full computations are not required, but any limit you rely on must be stated.

    1. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{n^2}{3n^2+1}}$ (alternative offered: the Ratio Test)
    2. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{n!}{n^n}}$ (alternative offered: the Root Test)
    3. ${\displaystyle {\sum }_{n=2}^{\infty }\frac{1}{n\sqrt{\ln n}}}$ (alternative offered: Direct Comparison with $\sum \frac{1}{n}$)
    4. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{(-1{)}^{n+1}}{2n+1}}$ (alternative offered: testing $\sum |a_n|$)
    5. ${\displaystyle {\sum }_{n=1}^{\infty }\frac{n^2+3}{n^5+n^2+1}}$ (alternative offered: the Ratio Test)
    6. ${\displaystyle {\sum }_{n=1}^{\infty }{\left(\frac{n}{n+3}\right)}^{n^2}}$ (alternative offered: the Ratio Test)
11. Q11Synthesis — drawing on several topics in this set

    Let ${\displaystyle S={\sum }_{n=1}^{\infty }(\frac{1}{n(n+2)}+\frac{5}{3^n})}$.

    1. Show that each of the two pieces converges, naming its type, and find the exact value of $S$.
    2. Splitting a series into two like that is legitimate here but not in general. Show that ${\displaystyle {\sum }_{n=1}^{\infty }(\frac{1}{n}-\frac{1}{n+1})}$ converges while neither piece does, and state the condition under which the linearity rule $\sum (a_n+b_n)=\sum a_n+\sum b_n$ may be used.

The 7 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

A sequence first, because every series is judged by one

Q1 asks which of four sequences converge, and where one does not, what the terms do instead. That second clause is the whole point. "Diverges" is not a synonym for "goes to infinity" — a sequence whose odd terms head one way and whose even terms head another diverges by oscillating, with every term staying bounded. Naming the behaviour is part of the answer, and the distinction returns the moment the alternating series arrive.

The sum of a series is the limit of its partial sums — that is the definition

Q2 is built to make that definition unavoidable: you are handed the partial sum S(n) of a filter-rinsing process in closed form and asked to recover the individual term a(n) from it, then to sum the series. The recovery is a(n) = S(n) − S(n−1), with the first index checked separately because there is no S(0) to subtract, and the sum is then a limit you already know how to take. Nothing here is a "test": the sum is computed, because the partial sum was available.

Q4 is the other half of that. Geometric and telescoping series are the two shapes where the partial sum can be written down in closed form, so they are the two where an exact sum is reachable — a geometric one because the ratio of consecutive terms is constant, a telescoping one because partial fractions split each term into pieces that cancel against their neighbours. Everything after Q4 is a test, and a test answers a yes-or-no question about convergence. It does not hand you the sum, so a question that asks for an exact value is telling you it is one of these two shapes.

Choosing the test: what the last sheet is really examining

Q10 gives six series, each with a named alternative test, and asks three things of each: the one test you would apply and why the shape of the general term invites it, why the stated alternative fails or is inconclusive, and the conclusion. Full computations are not wanted. That is not a shortcut — it is the actual skill, and it is what the ministerial competency means by pertinent application: not that you can run a test, but that you chose the one that settles this series and can say why the other one does not.

Notice what the alternative in each Q10 item is doing, because they are not all the same kind of failure. Sometimes it is a test that is inconclusive — it applies, it runs, and it returns no verdict. Sometimes it is a test that is inapplicable, because a hypothesis fails and it may not be run at all. Sometimes it is a comparison whose inequality points the useless way, which is the subtlest of the three and the one worth writing out in full. Three different sentences; "the Ratio Test does not work" is none of them, and it is the answer that loses the mark.

The Divergence Test, and the one direction it runs

Q3 also asks you to produce the general term a(n) from the first few written-out terms, which is a small skill with an outsized failure rate. Handle the three ingredients separately: the numerators as a pattern, the denominators as a pattern, and — if the signs alternate — a factor of (−1) raised to a power chosen so that n = 1 gives the sign the series actually starts with. Check the formula at n = 1 and n = 2 before using it. A general term that is off by one index makes every test after it technically about a different series.

The Integral Test: three hypotheses, and the integral is not the sum

Q5 says it outright — state and check the hypotheses on the companion function before integrating. There are three, and each is its own sentence: f is continuous on the interval, positive there, and decreasing there. Decreasing is the one that takes work, and the way to establish it is to differentiate f and check the sign, not to assert it from the look of the formula. The interval starts where the series starts — [1, ∞) or [2, ∞), whichever the summation index says — and if the front of the series misbehaves you may start later instead, because dropping finitely many terms cannot change whether a series converges.

Comparison: the inequality has to point the useful way

Q6 asks for the test used, the comparison series, why that series converges or diverges, and the inequality or limit that justifies the transfer. All four parts are the answer. The Direct Comparison Test is the sharper instrument when a clean inequality is available — dropping a positive term from a denominator makes the fraction larger, and that is usually where the inequality comes from — but it only ever transfers in two directions:

For the Limit Comparison Test the comparison series is read off the leading behaviour: keep the highest power in the numerator and the highest in the denominator, cancel, and compare against that p-series. The hypothesis to state is that the limit of the ratio is finite and positive — a limit of 0 or of ∞ belongs to a different, one-sided form of the test with a different conclusion, and quoting the standard form there is a real error rather than a shortcut. And once you have the comparison series you still owe the reason it converges or diverges: p > 1 or p ≤ 1, named. The harmonic series is where p = 1 sits, and its divergence is what this whole family of comparisons is anchored on.

Ratio and root: the two tests that read the term's construction

Q7 asks for the limit L and the conclusion, with the choice of test left to you and the shape of the term deciding it. The mechanics are worth rehearsing, because they are where the algebra bites. In a ratio of factorials, expand the larger one down to the smaller — (n+1)! = (n+1)·n!, and (2n+2)! = (2n+2)(2n+1)(2n)! — and cancel before taking any limit; a limit attempted on unexpanded factorials is unreadable. In a root, the nth root of a bracket raised to the power n is just the bracket, which is the entire reason that test exists.

Both tests conclude the same way: L < 1 gives convergence, L > 1 gives divergence, L = 1 gives nothing. Both also test absolute convergence, since both work with magnitudes — which is why the Ratio Test can be applied directly to a series whose terms alternate, and why a convergence verdict from it is automatically the stronger one.

Alternating series: two hypotheses, and "eventually" is doing real work

Q8(a) asks you to verify both hypotheses of the Alternating Series Test on a term of the form (−1) to the n+1 times b(n), and then to comment on what happens between the first two indices. That comment is the question. The terms b(n) must be positive, must tend to 0, and must decrease — but only eventually. A term that rises before it starts falling breaks nothing, because discarding a finite initial block changes no convergence question. Establish the decrease the same way as in the Integral Test: differentiate the companion function, find where the derivative turns negative, and say the decrease holds from that index onwards.

Absolute against conditional: two questions, asked in order

Q9 asks for a classification — absolutely convergent, conditionally convergent, or divergent — with the test named for each series it is applied to. The procedure is fixed, and following it in order saves work:

The asymmetry between the two directions is the fact to carry away. Divergence of the absolute series never implies divergence of the original — that is exactly what conditional convergence is. So a failed absolute test is a reason to keep working, never a verdict.

Splitting a series is a privilege, not a rule

Q11 closes the set with a series whose general term is a sum of two pieces, one telescoping and one geometric, and asks for the exact value. Splitting it is legitimate here — and part (b) then shows a series where the same move produces nonsense, because both pieces diverge while the series formed from their difference converges perfectly well.

Getting the most out of it

Write the general term before you do anything else

Half the errors in this set are made before a test is even chosen. Get a(n) in closed form, check it at the first two indices, and note where the summation actually starts — a series beginning at n = 2 rather than n = 1 changes the interval in the Integral Test and can decide whether a logarithm is defined at all. Only then look at the shape.

Name the test, then verify its hypotheses before you run it

Every test on this sheet is a theorem with conditions: positive terms for the comparisons, positive-continuous-decreasing for the integral, eventually decreasing terms tending to 0 for the alternating series test. A CEGEP marking guide gives those checks their own marks, and a correct verdict reached without them is a guess that happened to land. Write them as sentences, one per line.

Say which of "inconclusive" and "inapplicable" you mean

A test can return no information — the Ratio Test at L = 1, a comparison pointing the useless way — or it can be unavailable because one of its hypotheses fails. Those are different findings, and the closing sheet asks you to tell them apart. "It does not work" is neither of them.

Never conclude convergence from terms that shrink

If you take one habit from this worksheet, take this one. The Divergence Test can prove that a series diverges and can never prove that one converges. Every time you write "the terms tend to 0", the next words have to be "so the Divergence Test is inconclusive" — and then you go and choose a real test.

Common questions