# Thread: convergence divergence question quick and easy :D

1. ## convergence divergence question quick and easy :D

when taking the summation of $1/n$ from 1 to infinity,

what is the difference between saying that it diverges or converges to 0?

I am kind of puzzled by this because, for example the summation from 1 to infinity of $ln(2x+3)/x$, what does this do?

does it converge to 0 after using l'hopitals rule, or does it follow the harmonic series rule and diverge?

2. Originally Posted by p00ndawg
when taking the summation of $1/n$ from 1 to infinity,

what is the difference between saying that it diverges or converges to 0?

I am kind of puzzled by this because, for example the summation from 1 to infinity of $ln(2x+3)/x$, what does this do?

does it converge to 0 after using l'hopitals rule, or does it follow the harmonic series rule and diverge?
No...it does not ...the first one diverges because it is a divergent P-series..the second one diverges by a comparison to $\frac{\ln(x)}{x}$ which diverges by the integral test...if you need A LOT of help look here http://www.mathhelpforum.com/math-he...-tutorial.html

3. Originally Posted by Mathstud28
No...it does not ...the first one diverges because it is a divergent P-series..the second one diverges by a comparison to $\frac{\ln(x)}{x}$ which diverges by the integral test...if you need A LOT of help look here http://www.mathhelpforum.com/math-he...-tutorial.html
naw I dont need a lot of help, but I got a test tomorrow and just have a few holes i needed some filling on.

you say $1/n$ is divergent because of the p series, but isnt the series for it being greater than or less than 1? if its a big proof just reply with because I said so, ill understand.

but, so are you saying that if a question or function reduces through some kind of comparison or test, like the integral or nth term test, then if it does reduce to $1/n$ it would actually converge instead of diverge?

oh and im sorry the equation is $ln(2x^3 +1)/x$

4. Originally Posted by p00ndawg
when taking the summation of $1/n$ from 1 to infinity,

what is the difference between saying that it diverges or converges to 0?

I am kind of puzzled by this because, for example the summation from 1 to infinity of $ln(2x+3)/x$, what does this do?

does it converge to 0 after using l'hopitals rule, or does it follow the harmonic series rule and diverge?

The difference between a series diverging or converging to zero is a big difference. When a series diverges, that means the series has no sum. If a series converges to zero, that means that the sum of the series exists and is equal to zero.

The series $\sum\frac{\ln(2x+3)}{x}$ diverges; although L'Hospital's Rule can tell you that the limit of the terms is zero, this does not imply that the series converges. If the limit of the terms (as n approaches infinity) is not zero, then you can conclude the series diverges.

You show that $\sum\frac{\ln(2x+3)}{x}$ diverges by comparison with the harmonic series.

5. Originally Posted by p00ndawg
naw I dont need a lot of help, but I got a test tomorrow and just have a few holes i needed some filling on.

you say $1/n$ is divergent because of the p series, but isnt the series for it being greater than or less than 1? if its a big proof just reply with because I said so, ill understand.

but, so are you saying that if a question or function reduces through some kind of comparison or test, like the integral or nth term test, then if it does reduce to $1/n$ it would actually converge instead of diverge?
another way is this...since it decreases as n gets bigger and all the terms are positive we can apply the integral test $\int_1^{\infty}\frac{1}{n}dn=\ln(n)\bigg|_1^{\inft y}=\infty-0$ therefore it is divergent....and if a p-series has exponents 0<x<1 it diverges

6. Originally Posted by Mathstud28
another way is this...since it decreases as n gets bigger and all the terms are positive we can apply the integral test $\int_1^{\infty}\frac{1}{n}dn=\ln(n)\bigg_|1^{\inft y}=\infty-0$ therefore it is divergent....and if a p-series has exponents 0<x<1 it diverges
hey sorry about this but the equation was $ln(2x^3+1)/x$

7. Originally Posted by icemanfan

The difference between a series diverging or converging to zero is a big difference. When a series diverges, that means the series has no sum. If a series converges to zero, that means that the sum of the series exists and is equal to zero.

The series $\sum\frac{\ln(2x+3)}{x}$ diverges; although L'Hospital's Rule can tell you that the limit of the terms is zero, this does not imply that the series converges. If the limit of the terms (as n approaches infinity) is not zero, then you can conclude the series diverges.

You show that $\sum\frac{\ln(2x+3)}{x}$ diverges by comparison with the harmonic series.
does it change if the equation is $ln(2x^3+3/x)$?

8. Originally Posted by p00ndawg
hey sorry about this but the equation was $ln(2x^3+1)/x$
Originally Posted by p00ndawg
does it change if the equation is $ln(2x^3+3/x)$?
The first one still diverges by comparison to that $\frac{\ln(x)}{x}$ and the second one diverges since $\lim_{x\to\infty}\ln\bigg(2x^2+\frac{3}{x}\bigg)=\ infty\ne{0}$...which is the n-th term test

9. Originally Posted by p00ndawg
does it change if the equation is $ln(2x^3+3/x)$?
That series diverges even faster. And in that case, you can use the nth term test to prove it.