Infinite Series ?COnverge or Diverge

Click For Summary

Homework Help Overview

The discussion revolves around the convergence or divergence of two infinite series: ∑(infinity, k=1) 5k^(-3/2) and ∑(infinity, k=1) 1/(k+3). Participants are exploring the criteria for convergence in the context of infinite series.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest using the integral test for the first series and a comparison test with the harmonic series for the second. Others express a desire to avoid certain tests and seek alternative methods. There is also a discussion about the Test for Divergence and its applicability to both series.

Discussion Status

The conversation is ongoing, with participants sharing insights about different tests for convergence and divergence. There is a mix of agreement and confusion regarding the application of these tests, particularly the Test for Divergence. Some participants are questioning the assumptions underlying their reasoning.

Contextual Notes

One participant notes a restriction on using specific tests, which may affect the approaches discussed. Additionally, there are mentions of LaTeX formatting challenges related to mathematical expressions.

SAT2400
Messages
68
Reaction score
0
Infinite Series!??COnverge or Diverge

Homework Statement


1.
∑(infinity, k=1) 5k^(-3/2)

2.
∑(infinity, k=1) 1/(k+3)

Homework Equations


converge or diverge

The Attempt at a Solution


1. converges p=3/2... 5/(infin)=> 0
2. diverges p=1
I still don't get why 2 diverges? 1 converges b/c it gets to 0??
 
Physics news on Phys.org


use integral test for 1

for 2 try and find a comparison test with a harmonic series(1/k)
 


Umm, I am not supposed to use those tests...

Is there any other method??
 


Can't you just use the test for divergence for #2?

EDIT: I think it works for both #1 and #2.
 


could you please show me?thanks
 


The best that I can do is tell you what the Test for Divergence is.

The Test for Divergence states that:

If lim _{x\rightarrow\infty} a_{n} \neq 0, \sum a_{n} diverges.

P.S. My LaTeX is weak, could someone tell me how to add spaces?
 


SAT2400 said:
Umm, I am not supposed to use those tests...

Is there any other method??

what can you use? i can't read your question/notes/teachers mind ;)
 


mg0stisha said:
The best that I can do is tell you what the Test for Divergence is.

The Test for Divergence states that:

If lim _{x\rightarrow\infty} a_{n} \neq 0, \sum a_{n} diverges.

i think both the terms go to zero in the limit? so it doesn't show divergence?

mg0stisha said:
P.S. My LaTeX is weak, could someone tell me how to add spaces?
as for latex, just found out myself (from other posts)-, you can also add the slash to make the limit show correctly, have a look at these:
single space "\"
\lim _{x\rightarrow\infty} a_{n} \neq 0, \ \sum a_{n} \ diverges.

multispace 0.5 inches "\hspace{0.5 in}"
\lim _{x\rightarrow\infty} a_{n} \neq 0, \hspace{0.5 in} \sum a_{n} \ diverges.

using itex (inline tex) and splitting into 2 parts
\lim _{x\rightarrow\infty} a_{n} \neq 0, \sum a_{n} \ diverges.
 
Last edited:


mg0stisha said:
Can't you just use the test for divergence for #2?

EDIT: I think it works for both #1 and #2.

what are you talking about? #1 converges and #2 does diverge but you're not going to show it using the divergence test since lim_{k\rightarrow \infty} \frac{1}{(k+3)} = 0

to op:

for #1 it's a p series
for #2 show that it's bigger than \sum \frac{1}{5k} and then show that that diverges
 
  • #10


Wow, I apologize to the OP. Guess I should stop doing math at 4 am and just go to bed! Sorry for any confusion, I definitely see my blindingly obvious mistakes now.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
7
Views
2K