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Practice problems on sequences and series

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Sequences and Series Quiz 1

"},"excerpt":{"rendered":"

Practice problems on sequences and series

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":15,"niveau_difficulte":"$undefined","categorie":"Sequences and Series","nombre_questions":10,"score_minimum":70,"ordre_questions":"Fixe","questions":[{"id":"cmpz9r6bf05p8jgllu37qjugv","texte_question":"

Determine whether or not the sequence $$\\left\\{\\frac{4n^3+3n-2}{7n^3-5n^2+1}\\right\\}$$ is convergent or divergent.

","type_question":"QCM","explication":"

For a rational sequence with the same highest power in numerator and denominator, the limit is the ratio of the leading coefficients: $$\\lim_{n\\to\\infty}\\frac{4n^3+3n-2}{7n^3-5n^2+1}=\\frac{4}{7}$$ So the sequence converges to $$\\frac47$$. Final answer: Converges to $$\\frac47$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Converges to $$0$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Converges to $$\\frac47$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"Diverges to infinity","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Diverges","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6bh05pejglltsg2wfnk","texte_question":"

Find the sum of the series $$\\sum_{n=1}^{\\infty}\\left(\\frac{1}{n+2}-\\frac{1}{n+4}\\right)$$.

","type_question":"QCM","explication":"

This is a telescoping series. Write out the first terms: $$\\left(\\frac13-\\frac15\\right)+\\left(\\frac14-\\frac16\\right)+\\left(\\frac15-\\frac17\\right)+\\left(\\frac16-\\frac18\\right)+\\cdots$$ Most terms cancel, leaving $$\\frac13+\\frac14=\\frac{7}{12}$$ Final answer: $$\\frac{7}{12}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac12$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac34$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac{7}{12}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\frac23$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6bl05pkjgllrhww0ko9","texte_question":"

Determine if the sequence $$\\left\\{\\frac{\\ln n}{n}\\right\\}$$ converges.

","type_question":"QCM","explication":"

As $$n\\to\\infty$$, the denominator grows faster than the numerator. Also, using the corresponding continuous limit, $$\\lim_{x\\to\\infty}\\frac{\\ln x}{x}=0$$ So the sequence converges to $$0$$. Final answer: Converges to $$0$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Converges to $$0$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"Converges to $$1$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Diverges","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Oscillates","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6bn05pqjgllyfz2w83b","texte_question":"

Find the $$n$$th term and determine convergence for the sequence $$3,\\frac32,\\frac34,\\frac38,\\frac{3}{16},\\ldots$$.

","type_question":"QCM","explication":"

Each term is obtained by multiplying by $$\\frac12$$, so this is a geometric sequence with first term $$3$$ and ratio $$\\frac12$$. Thus $$a_n=3\\left(\\frac12\\right)^{n-1}=\\frac{3}{2^{n-1}}$$ As $$n\\to\\infty$$, $$\\frac{3}{2^{n-1}}\\to0$$ Final answer: $$a_n=\\frac{3}{2^{n-1}}$$, converges to $$0$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$a_n=\\frac{3}{2^n}$$, converges to $$0$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$a_n=\\frac{3}{n}$$, converges to $$0$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$a_n=\\frac{3}{2n}$$, diverges","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$a_n=\\frac{3}{2^{n-1}}$$, converges to $$0$$","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"}]},{"id":"cmpz9r6bq05pwjglluvlng7da","texte_question":"

Use the nth-term divergence test to determine whether the series $$\\sum_{n=1}^{\\infty}\\frac{2n^3+1}{5n^3-2n+4}$$ converges.

","type_question":"QCM","explication":"

For the nth-term test, compute the limit of the terms: $$\\lim_{n\\to\\infty}\\frac{2n^3+1}{5n^3-2n+4}=\\frac25\\neq0$$ Since the terms do not approach $$0$$, the series diverges. Final answer: Diverges

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Converges","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Diverges","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"Conditionally converges","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"Cannot determine","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6bt05q2jgll870p8bg8","texte_question":"

What is the sum of $$\\sum_{n=1}^{\\infty}\\left(\\frac{1}{n+2}-\\frac{1}{n+5}\\right)$$?

","type_question":"QCM","explication":"

This is telescoping. Write the first terms: $$\\left(\\frac13-\\frac16\\right)+\\left(\\frac14-\\frac17\\right)+\\left(\\frac15-\\frac18\\right)+\\cdots$$ Everything cancels except the first three positive terms: $$\\frac13+\\frac14+\\frac15$$ Find the sum: $$\\frac13+\\frac14+\\frac15=\\frac{20+15+12}{60}=\\frac{47}{60}$$ Final answer: $$\\frac{47}{60}$$ Note: the source answer choices did not include the correct sum, so the answer set was corrected.

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$\\frac{47}{60}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$\\frac{7}{12}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac{11}{12}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$\\frac56$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6bw05q8jgllqaex839j","texte_question":"

Determine whether or not the series converges $$\\sum_{n=1}^{\\infty}\\frac{1}{\\sqrt[4]{n^6+2}}$$

","type_question":"QCM","explication":"

For large $$n$$: $$\\sqrt[4]{n^6+2}\\sim n^{6/4}=n^{3/2}$$ So the series behaves like $$\\sum \\frac{1}{n^{3/2}}$$ Since $$p=\\frac32>1$$, the p-series converges. Final answer: Converges by comparison test

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Converges by comparison test","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"Diverges by p-series test","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Converges conditionally","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Diverges by ratio test","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6by05qejgllzd67i6ur","texte_question":"

Determine whether or not the series converges $$\\sum_{n=1}^{\\infty}\\frac{n^2}{n!}$$

","type_question":"QCM","explication":"

Use ratio test: $$\\lim_{n\\to\\infty}\\left|\\frac{(n+1)^2/(n+1)!}{n^2/n!}\\right| =\\lim_{n\\to\\infty}\\frac{(n+1)^2}{(n+1)n^2} =\\lim_{n\\to\\infty}\\frac{n+1}{n^2}=0$$ Since limit is 0 < 1, series converges absolutely. Final answer: Converges absolutely

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Converges absolutely","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Diverges","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Converges conditionally","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Oscillates","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6c105qkjgllgvyyxy0y","texte_question":"

Determine whether or not the series converges $$\\sum_{n=1}^{\\infty}\\sqrt{\\frac{5n+2}{n^6+1}}$$

","type_question":"QCM","explication":"

For large $$n$$: $$\\sqrt{\\frac{5n+2}{n^6+1}}\\sim\\sqrt{\\frac{5n}{n^6}}=\\sqrt{\\frac5{n^5}}=\\frac{\\sqrt5}{n^{5/2}}$$ Since $$p=\\frac52>1$$, the series converges by comparison. Final answer: Converges by comparison test

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Converges by comparison test","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Diverges by p-series","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

Converges conditionally

","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Diverges by nth-term test","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6c305qqjglleku07j6f","texte_question":"

Determine whether or not the series converges $$\\sum_{n=1}^{\\infty}\\frac{4^n+2}{3^n}$$

","type_question":"QCM","explication":"

Rewrite: $$\\sum\\left(\\frac{4^n}{3^n}+\\frac{2}{3^n}\\right) =\\sum\\left(\\left(\\frac43\\right)^n+2\\left(\\frac13\\right)^n\\right)$$ Since $$\\left(\\frac43\\right)^n$$ diverges, the series diverges. Final answer: Diverges

Converges conditionally

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

Alternating convergence

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]}]},"categories":["Sequences and Series"],"date":"2026-06-04T09:04:29.112Z","modified":"2026-06-04T09:04:29.112Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L18",null,{"targetType":"quiz","targetSlug":"ap-calculus-bc-sequences-series-quiz-1"}]}]]}]}]]

Quiz 1

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