Tricky calculus question - i've got problems

  • Thread starter Thread starter herraotic
  • Start date Start date
  • Tags Tags
    Calculus
Click For Summary
SUMMARY

The discussion centers on proving the divergence of the series defined by the general term min(an, 1/n), where (an) is a sequence of positive real numbers that is decreasing and has an infinite sum. Participants emphasize that demonstrating the monotonic nature of the sequence is crucial. The key conclusion is that since the sequence (an) is decreasing and the terms are bounded below by 1/n, the series diverges as well.

PREREQUISITES
  • Understanding of sequences and series in calculus
  • Knowledge of monotonic sequences
  • Familiarity with the concept of divergence in mathematical analysis
  • Basic proficiency in using the min function in mathematical expressions
NEXT STEPS
  • Study the properties of monotonic sequences in calculus
  • Learn about convergence and divergence criteria for series
  • Explore the application of the min function in mathematical proofs
  • Investigate examples of divergent series involving sequences
USEFUL FOR

Students of calculus, mathematicians interested in series and sequences, and educators looking for examples of divergence proofs.

herraotic
Messages
4
Reaction score
0
Let (an) be a sequence of positive real numbers, decreasing and the sum of whose terms is infinite. Prove that the series whose general term is min(an, 1/n) is also divergent.

I'm sorry but I'm no where with it. Could someone tell me if this is too difficult?

Thanks!
 
Physics news on Phys.org
herraotic said:
Let (an) be a sequence of positive real numbers, decreasing and the sum of whose terms is infinite. Prove that the series whose general term is min(an, 1/n) is also divergent.

I'm sorry but I'm no where with it. Could someone tell me if this is too difficult?

Thanks!

I think that all you have to show here is that it is a monotonic decreasing sequence

nth term -> min (an or 1/n)

for nth+1 term -> min (an+1 or 1/(n+1))

the 'min' function says that you choose 1/n or something lower than 1/n

so you can tell for sure that

an+1 < an
 
I'm still stumped. Why don't you let me off and show the solution :D :D ;D
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
32
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
8
Views
3K
Replies
29
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K