Comparison test of infinite series

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SUMMARY

The discussion centers on the divergence of the series represented by the harmonic series, specifically ##\sum_{n = 1}^\infty \frac{1}{n}##. Participants clarify that while the sequence of terms ##\{a_n\}## converges to zero, the series itself diverges, emphasizing the importance of evaluating the sequence of partial sums ##S_n = \sum_{k = 1}^n a_k##. The Nth Term Test for Divergence is highlighted as a critical tool, where if ##\lim_{n \to \infty} a_n \ne 0##, the series diverges. Misinterpretations regarding the notation ##1/3n## are also addressed, clarifying that it denotes ##\frac{1}{3}n## rather than ##\frac{1}{3n}##.

PREREQUISITES
  • Understanding of infinite series and convergence concepts
  • Familiarity with the Nth Term Test for Divergence
  • Knowledge of the comparison test in calculus
  • Basic proficiency in limits and notation in calculus
NEXT STEPS
  • Study the properties of the harmonic series and its divergence
  • Learn about the comparison test for series convergence
  • Explore the implications of the Nth Term Test for Divergence
  • Review limit notation and its proper usage in calculus
USEFUL FOR

Calculus students, educators, and anyone seeking to deepen their understanding of infinite series and convergence tests in mathematical analysis.

isukatphysics69
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Homework Statement



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The Attempt at a Solution


So the book is saying that this series diverges, i have learned my lesson and have stopped doubting the authors of this book but i don't understand how this series diverges. ok i can use the comparison test using 1/3n and 1/3n converges, since 1/(3n+1) is less than 1/3n i don't see how it will not also converge? plug in infinity to 1/3n and you get 0 which converges
 

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You are not worried about the terms themselves, you are worried about the sum.
isukatphysics69 said:
and 1/3n converges
...
plug in infinity to 1/3n and you get 0 which converges
No. ##\displaystyle {\sum{1\over n}}## does NOT converge, so 1/3n doesn't either.
 
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I was mistaken you are right thank you
 
i don't know how i will solve the Riemann hypothesis if i cannot even solve this stuff
 
Cheer up. :biggrin: Insight grows. From exercises that you can do easily, you don't learn fast enough. And insight grows.
And you don't have to solve the thing, just build some understanding.
 
isukatphysics69 said:
i can use the comparison test using 1/3n and 1/3n converges

BvU said:
so 1/3n doesn't either
Per the rules of PEMDAS, 1/3n is normally taken to mean ##\frac 1 3 n##, not ##\frac 1 {3n}##. Written inline this fraction should be 1/(3n) so as to not be ambiguous.
 
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isukatphysics69 said:
So the book is saying that this series diverges, i have learned my lesson and have stopped doubting the authors of this book but i don't understand how this series diverges. ok i can use the comparison test using 1/3n and 1/3n converges, since 1/(3n+1) is less than 1/3n i don't see how it will not also converge? plug in infinity to 1/3n and you get 0 which converges
The topic of infinite series has some subtleties that cause confusion in many calculus students.

If a series ##\sum_{n = 1}^\infty a_n## to converges, it's irrelevant that the sequence of terms, ##\{a_n\}## converges. What is relevant is wheter the sequence of partial sums converges or diverges. In other words, whether ##\lim_{n \to \infty} S_n## converges. Here ##S_n = \sum_{k = 1}^n a_k = a_1 + a_2 + \dots + a_n##

The sequece of terms in a series can converge to zero, without the series itself converging. This means that most of the time, evaluating the limit ##\lim_{n \to \infty} a_n## is a wasted effort. The only time this is useful is when ##\lim_{n \to \infty} a_n \ne 0##, from which we can conclude by the Nth Term Test for Divergence, that the series definitely diverges. We cannot conclude anything at all if this same limit happens to be zero.
For example, it's well known that the harmonic series, ##\sum_{n = 1}^\infty \frac 1 n## diverges, even though ##\lim_{n \to \infty}\frac 1 n = 0##

For the series ##\sum_{n = 1}^\infty n##, since ##\lim_{n \to \infty}n = \infty \ne 0##, we can conclude that this series diverges, by the Nth Term Test for Divergence.

isukatphysics69 said:
plug in infinity to 1/3n and you get 0 which converges
Two things here:
First, we don't "plug in infinity" since infinity is not a number we can use in any arithmetic or algebraic expression. Instead, we look at a limit.
Second, as I mentioned in a previous post, 1/3n means ##\frac 1 3 n##, so ##\lim_{n \to \infty} \frac 1 3 n = \frac 1 3 \lim_{n \to \infty}n = \infty##.
 

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