Is \sum (\sqrt{k+1} - \sqrt{k})(\ln{k+1}-\ln{k}) convergent or divergent?

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SUMMARY

The series \(\sum_{k=1}^\infty (\sqrt{k+1} - \sqrt{k})(\ln{k+1}-\ln{k})\) is under analysis for convergence or divergence. The discussion highlights the transformation of the term \(\sqrt{k+1} - \sqrt{k}\) into \(\frac{1}{\sqrt{k} + \sqrt{k+1}}\) and the combination of logarithmic terms. A key insight provided is the inequality \(\frac{1}{x} \geq \ln(1 + \frac{1}{x})\) for all positive \(x\), which can be utilized to establish convergence properties.

PREREQUISITES
  • Understanding of series convergence tests
  • Familiarity with logarithmic properties
  • Knowledge of limits and inequalities in calculus
  • Basic manipulation of square roots and algebraic expressions
NEXT STEPS
  • Study the comparison test for series convergence
  • Learn about the properties of logarithmic functions
  • Explore the application of the squeeze theorem in series
  • Investigate the behavior of \(\sqrt{k}\) and its limits as \(k\) approaches infinity
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Mathematicians, calculus students, and anyone interested in series analysis and convergence tests will benefit from this discussion.

Hydr0matic
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<br /> \sum_{k=1}^\infty (\sqrt{k+1} - \sqrt{k})(\ln{k+1}-\ln{k})<br />

How do I go about finding out if it's convergent or divergent ?
 
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Hydr0matic said:
<br /> \sum_{k=1}^\infty (\sqrt{k+1} - \sqrt{k})(\ln{k+1}-\ln{k})<br />

How do I go about finding out if it's convergent or divergent ?
Do you mean:
<br /> \sum_{k=1}^\infty (\sqrt{k+1} - \sqrt{k})(\ln{(k+1)}-\ln{k})<br />?
 
yes.. thnx.
 
Hint: sqrt(k+1)-sqrt(k) = {sqrt(k)+sqrt(k+1)}^{-1}
and you can put the logs together.
 
Got it. Thnx matt.
 
after combining the logs, try to prove that
\frac{1}{x} \geq \ln (1 + \frac{1}{x})
for all positive x

edit:
oops i missed the last reply while typing mine sorry
 

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