Does Divergence of a Series Imply Divergence of Its Absolute Values?

Click For Summary
SUMMARY

The discussion centers on proving that if the series Σa_n diverges, then the series Σ|a_n| also diverges. The Comparison Test is referenced, highlighting that it requires both sequences to be non-negative, which is not applicable in this case. The Cauchy criterion is introduced as a potential method for proof, emphasizing the relationship between the absolute sum and the original series. The conclusion is that the absolute values of the terms must also diverge if the original series does.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with the Comparison Test in series
  • Knowledge of the Cauchy criterion for series
  • Basic concepts of absolute values in mathematical analysis
NEXT STEPS
  • Study the Cauchy criterion in detail for series convergence
  • Explore the properties of absolute convergence versus conditional convergence
  • Learn about the implications of the Comparison Test in various contexts
  • Investigate examples of divergent series and their absolute values
USEFUL FOR

Mathematics students, educators, and anyone studying series convergence and divergence in real analysis.

musashi1029
Messages
8
Reaction score
0

Homework Statement



Show that following statement is true:
If Σa_n diverges, then Σ|a_n| diverges as well.

Homework Equations



Comparison Test:
If 0 ≤ a_n ≤ b_n for all n ≥ 1, and if Σa_n diverges, then Σb_n diverges as well.

The Attempt at a Solution



I tried to prove the statement by using the Comparison Test with a_n = a_n and b_n = |a_n|, but the condition for the Comparison Test is that both sequences must be greater than or equal to zero, which is not true for this problem. I would like to know if using the Comparison Test is a right approach to prove this statement.

Thank you in advance.
 
Physics news on Phys.org
Think of the Cauchy criterion; that is, \sum a_n converges if and only if for all \epsilon >0 there exists an integer N such that \left| \sum_{k=n}^ma_k \right| \leq \epsilon if m\geq n\geq N.

Now think of what you have to prove and the fact that

\left| \sum_{k=n}^ma_k \right| \leq \sum_{k=n}^m |a_k|
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
29
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K