basic infinite series

[logistic_guy]

here is the question

Does the infinite series \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\) converge or diverge?


my attemb
this series is famous for divergence
am i correct?

 

[fresh_42]

Yes. Do you have an idea of how to prove it?

 

[logistic_guy]

thank

Yes.

the question don't ask to show if the series converge or diverge
don't this mean my answer is complete?

Do you have an idea of how to prove it?

i've not only idea but many ideas
i know there is some tests to use
the problem i'm still lack the skills of how to chose the correct test

if i randomly chose the comparison test
i know \(\displaystyle \frac{1}{n} > \frac{1}{n^2}\)
i also know \(\displaystyle \frac{1}{n^2}\) converge
i don't know how to use this idea to say \(\displaystyle \frac{1}{n}\) diverge

 

[fresh_42]

the question don't ask to show if the series converge or diverge
don't this mean my answer is complete?

I would say no since a mathematical answer without proof is problematic. It's not legal science or biology where facts are asked.

i've not only idea but many ideas
i know there is some tests to use
the problem i'm still lack the skills of how to chose the correct test

The oldest proof I've read about is from the 14th century! It is far easier than any criteria. We are asked about the value of
[math] 1+\dfrac{1}{2} + \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\ldots [/math]and we can group the terms:
[math] 1+\dfrac{1}{2} + \left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}\right)+\ldots [/math]Now every sum in the parentheses is greater than [imath] 1/2 [/imath] and thus they sum up to infinity.

All convergence criteria come down to a direct comparison test if you look at their proofs: find a convergent series that is greater to prove convergence and a divergent series that is less for divergence. Here is a little article about it:

Series in Mathematics: From Zeno to Quantum Theory

Explore the profound impact of series in mathematics, tracing their historical significance from ancient mathematicians to quantum theory.

www.physicsforums.com

 

[khansaheb]

Do a Google search with the keywords - "integration test for convergence of a series"

Tell us - what did you find.

 

[logistic_guy]

The oldest proof I've read about is from the 14th century! It is far easier than any criteria. We are asked about the value of
[math] 1+\dfrac{1}{2} + \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\ldots [/math]and we can group the terms:
[math] 1+\dfrac{1}{2} + \left(\dfrac{1}{3}+\dfrac{1}{4}\right)+\left(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}\right)+\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{16}\right)+\ldots [/math]Now every sum in the parentheses is greater than [imath] 1/2 [/imath] and thus they sum up to infinity.

again you explain it nicely fresh_42
i think i understand this classical proof

All convergence criteria come down to a direct comparison test if you look at their proofs: find a convergent series that is greater to prove convergence and a divergent series that is less for divergence. Here is a little article about it:

Series in Mathematics: From Zeno to Quantum Theory

Explore the profound impact of series in mathematics, tracing their historical significance from ancient mathematicians to quantum theory.

www.physicsforums.com

i also think i understand this
you're saying if \(\displaystyle A < \frac{1}{n}\) and \(\displaystyle A\) diverge then \(\displaystyle \frac{1}{n}\) diverge
if \(\displaystyle A > \frac{1}{n}\) and \(\displaystyle A\) converge then \(\displaystyle \frac{1}{n}\) converge

is my analize correct?

Do a Google search with the keywords - "integration test for convergence of a series"

Tell us - what did you find.

thank khan

can you provide me a different method than what fresh_42 give me to proof the divergence of the op?

 

[fresh_42]

again you explain it nicely fresh_42
i think i understand this classical proof


i also think i understand this
you're saying if \(\displaystyle A < \frac{1}{n}\) and \(\displaystyle A\) diverge then \(\displaystyle \frac{1}{n}\) diverge
if \(\displaystyle A > \frac{1}{n}\) and \(\displaystyle A\) converge then \(\displaystyle \frac{1}{n}\) converge

is my analize correct?

Yes, theoretically, since [imath] \sum 1/n [/imath] does not converge so there is no series [imath] A=\sum a_n > \sum 1/n [/imath] that converges.

thank khan

can you provide me a different method than what fresh_42 give me to proof the divergence of the op?

Integral test for convergence - Wikipedia

en.wikipedia.org

 

[logistic_guy]

Yes, theoretically, since [imath] \sum 1/n [/imath] does not converge so there is no series [imath] A=\sum a_n > \sum 1/n [/imath] that converges.

Integral test for convergence - Wikipedia

en.wikipedia.org

i don't see why and how

\(\displaystyle \int_{1}^{\infty} \frac{1}{x} dx \leq \sum_{n=1}^{\infty}\frac{1}{n}\)

 

[topsquark]

i don't see why and how

\(\displaystyle \int_{1}^{\infty} \frac{1}{x} dx \leq \sum_{n=1}^{\infty}\frac{1}{n}\)

logicstic_guy, you can't be that bad at Calculus and have passed the course.

Click on the Wikipedia link. There is a graph on the right side of the screen. What does the integral [imath]\displaystyle \int_1^{\infty} \dfrac{1}{x} dx[/imath] mean? What does the sum of the areas of the rectangles mean?

-Dan

 

[fresh_42]

If English isn't your first language, maybe yours or a language you understand well enough is among the language versions on Wikipedia. Change the language. They are no 1:1 translations. Even I switch occasionally between different languages to see where something is explained best.

 

[logistic_guy]

logicstic_guy, you can't be that bad at Calculus and have passed the course.

thank topsquark very much
you're the first one to say nice words to me

Click on the Wikipedia link. There is a graph on the right side of the screen. What does the integral [imath]\displaystyle \int_1^{\infty} \dfrac{1}{x} dx[/imath] mean? What does the sum of the areas of the rectangles mean?

-Dan

i see
i'm not quite sure but i think the area under the curve is the integration and the rectangles area is the infinite sereis

is my analize correct?

If English isn't your first language, maybe yours or a language you understand well enough is among the language versions on Wikipedia. Change the language. They are no 1:1 translations. Even I switch occasionally between different languages to see where something is explained best.

thank fresh_42

 

[topsquark]

thank topsquark very much
you're the first one to say nice words to me


i see
i'm not quite sure but i think the area under the curve is the integration and the rectangles area is the infinite sereis

is my analize correct?


thank fresh_42

What does the infinite series represent?

-Dan

 

[logistic_guy]

What does the infinite series represent?

-Dan

i think it represent an approximation to the area under the curvei'm not sure

 

[topsquark]

i think it represent an approximation to the area under the curvei'm not sure

It's not an approximation, it is the area under the steps! How does this compare to the integral?

-Dan

 

[logistic_guy]

It's not an approximation, it is the area under the steps! How does this compare to the integral?

-Dan

i see now it's bigger. i think i get the idea of the comparison

thank i appreciaite it

\(\displaystyle \int_{1}^{\infty}\frac{1}{x} dx = \ln x\big|_{1}^{\infty} = \ln \infty - \ln 1 = \ln \infty - 0 = \ln \infty\)

how to solve \(\displaystyle \ln \infty\), there's no button in my calculator for the infinity

 

[topsquark]

i see now it's bigger. i think i get the idea of the comparison

thank i appreciaite it

\(\displaystyle \int_{1}^{\infty}\frac{1}{x} dx = \ln x\big|_{1}^{\infty} = \ln \infty - \ln 1 = \ln \infty - 0 = \ln \infty\)

how to solve \(\displaystyle \ln \infty\), there's no button in my calculator for the infinity

First, [imath]\infty[/imath] is not a number. You can't calculate with it.

Since you aren't bothering to go back and review:
[imath]\displaystyle \int_1^{\infty} \dfrac{1}{x} ~ dx = \lim_{M \to \infty} \int_1^M \dfrac{1}{x} ~ dx = \lim_{M \to \infty} ln(M) - ln(1) = \lim_{M \to \infty} ln(M)[/imath]

What is [imath]\displaystyle \lim_{M \to \infty} ln(M)[/imath]. And don't just give us a if you don't know. Play with the limit and find out!

-Dan

 

[logistic_guy]

[imath]\displaystyle \int_1^{\infty} \dfrac{1}{x} ~ dx = \lim_{M \to \infty} \int_1^M \dfrac{1}{x} ~ dx = \lim_{M \to \infty} ln(M) - ln(1) = \lim_{M \to \infty} ln(M)[/imath]

i think i'm understand your explanation

What is [imath]\displaystyle \lim_{M \to \infty} ln(M)[/imath]. And don't just give us a if you don't know. Play with the limit and find out!

\(\displaystyle \lim_{M \to \infty} \ln(M) = \ln(\infty)\)

month ago i remember how to draw the logarthm function
i forget nowand i've to look at it again

hold on topsquark i'll get back to you soon

 

[topsquark]

i think i'm understand your explanation


\(\displaystyle \lim_{M \to \infty} \ln(M) = \ln(\infty)\)

month ago i remember how to draw the logarthm function
i forget nowand i've to look at it again

hold on topsquark i'll get back to you soon

No. [imath]ln(\infty)[/imath] does not exist. It's meaningless. (That one isn't your fault. You've probably seen it in your classes or your textbooks.) You can't take the log of infinity because infinity isn't a number. It's the limit that you are looking at.

-Dan

 

[Steven G]

logicstic_guy, you can't be that bad at Calculus and have passed the course.

Unfortunately that is not always true. I knew of some adjuncts who passed the weakest of students just to have a good passing rate.
If you teach the material correctly, then your passing rate will be fine.

 

[logistic_guy]

Unfortunately that is not always true. I knew of some adjuncts who passed the weakest of students just to have a good passing rate.
If you teach the material correctly, then your passing rate will be fine.

thank steven very much for aiming to say i'm the weakest student

hold on topsquark i'll get back to you soon

i'm back

No. [imath]ln(\infty)[/imath] does not exist. It's meaningless. (That one isn't your fault. You've probably seen it in your classes or your textbooks.) You can't take the log of infinity because infinity isn't a number. It's the limit that you are looking at.

-Dan

i see the graph of the lograthm



i see it touch zero and go right
how this help me?