You're welcome! Glad it worked for you.

  • Context: Graduate 
  • Thread starter Thread starter mahmoud2011
  • Start date Start date
  • Tags Tags
    Inequality
Click For Summary
SUMMARY

The inequality \(0 \leq \sum_{k=0}^{n} \frac{1}{(k+1)^2 (n-k+1)} \leq \frac{1}{\sqrt{n+1}}\) is confirmed to be true for all natural numbers \(n\). Participants in the discussion suggest rewriting the term \(\frac{1}{(k+1)^2(n-k+1)}\) in a more manageable form, specifically as \(\frac{a}{k+1} + \frac{bk+c}{(k+1)^2} + \frac{d}{n-k+1}\). This transformation aids in the analysis and understanding of the inequality. The discussion emphasizes the importance of algebraic manipulation in proving mathematical inequalities.

PREREQUISITES
  • Understanding of mathematical inequalities
  • Familiarity with summation notation
  • Knowledge of algebraic manipulation techniques
  • Basic concepts of natural numbers
NEXT STEPS
  • Research algebraic techniques for manipulating summations
  • Study mathematical proofs involving inequalities
  • Explore advanced topics in series convergence
  • Learn about the properties of natural numbers in mathematical analysis
USEFUL FOR

Mathematicians, students studying advanced algebra, and anyone interested in the analysis of inequalities and summation techniques.

mahmoud2011
Messages
88
Reaction score
0
Is this inequality true ??

0\leq \sum_{k=0}^{n} \frac{1}{(k+1)^2 (n-k+1)} \leq \frac{1}{\sqrt{n+1}} for all natural numbers n

Is it true ??

Thanks
 
Mathematics news on Phys.org


Can you write \frac{1}{(k+1)^2(n-k+1)} differently? You want something on the form \frac{a}{k+1} + \frac{bk+c}{(k+1)^2} + \frac{d}{n-k+1} .
 


It really work thanks .
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K