Converging and Diverging Sequences Using Limits - Practice Problems
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- เผยแพร่เมื่อ 24 ก.ย. 2024
- This calculus video tutorial provides a basic introduction into converging and diverging sequences using limits. It explains how to write out the first four terms of a sequence and how to determine if a sequence converges or diverges by finding the limit of a sequence. If the limit exists and it equals to some constant L as n approaches infinity, then the sequence converges. If the limit does not exist or increases or decreases without bound, that is, to positive or negative infinity - then the sequence diverges. This video contains plenty of examples and practice problems of determining if a sequence is convergent or divergent using Lhopital's rule, the squeeze theorem, and properties of logarithms. It contains example problems with trigonometric functions such as sine and cosine, natural logarithms, square root functions, factorials, and exponential functions.
Improper Integrals: • Improper Integrals - C...
Converging & Diverging Sequences:
• Converging and Divergi...
Monotonic & Bounded Sequences:
• Monotonic Sequences an...
Absolute Value Theorem - Sequences:
• Absolute Value Theorem...
Squeeze Theorem - Sequences:
• Squeeze Theorem For Se...
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Geometric Series & Sequences:
• Geometric Series and G...
Introduction to Series - Convergence:
• Convergence and Diverg...
Divergence Test For Series:
• Divergence Test For Se...
Harmonic Series:
• Harmonic Series
Telescoping Series:
• Telescoping Series
__________________________________
Integral Test For Divergence:
• Calculus 2 - Integral ...
Remainder Estimate - Integral Test:
• Remainder Estimate For...
P-Series:
• P-series
Direct Comparison Test:
• Direct Comparison Test...
Limit Comparison Test:
• Limit Comparison Test
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Calculus Final Exam and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
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@@theadel8591 I feel like that'll be the case for us, except I bet most of it is going to be power series
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Yes it is cause the lim as n>infinity of ln(1+(1/n))/(1/n) is ln(1+0)/(0) which is undefined.
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Professor Organic Chemistry Tutor, thank you for using multiple well-known examples to explain/analyze Converging and Diverging Sequences in Calculus Two. In all Calculus textbooks, there is a whole chapter on Sequences and Series. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
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plugging in inf into (1+1/n)^n = (1+1/inf)^inf = (1+0)^inf = (1)^inf = 1^inf = undefined
this is why we had to solve it
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At 4:30 why do you multiply the top and bottom by 1/n? How do I recognize when to use this?
Did you find out the reason? I just started Pre calculus and would like to know thr proof for multiplying the top and bottom by 1/2
@@FerghusCameron
Just to take the n to an easier place. This can be solved by many ways, and you can keep the n and evaluate it to find the answer (if it's converging or diverging) but dividing by 1/n takes the n to 5/n which leads to the final answer easily (because of the rule 1/n=0).
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why do u use l'hopital rule insted of just substitute infinity in the equation lim n-> infinity n * sin(1/n)???
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why would you add the last step at 22:15? 1/(n+2) is just 1/infinity which is zero. Absolutely no need to multiply top and bottom to get the zero in the numerator
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Will give u a good idea about convergence nd divergence.
But formaly the idea of Convergence nd divergence is different.
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Wait so at 22:55 why do you multiply by 1/n instead of just evaluating the limit there??
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Should not we apply stolz-cesara instead of l’hospital when dealing with sequences? Please explain!
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What should you do if it goes to 0 with infinity but its limit for negative infinity is infinite? Does it till converge?