Finding the convergence of a parametric series

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SUMMARY

The discussion focuses on determining the convergence of a parametric series, specifically analyzing the behavior of the sequence terms as n approaches infinity. It is established that the terms tend to zero for certain values of α, with the conclusion that the sum converges when α is greater than or equal to 1. The participants explored various convergence tests, including the root criterion and the ratio test, but found them ineffective in this context. A transformation involving ##\sqrt[n]{n}## to ##e^{\frac{ln n}{n}}## was suggested, indicating a need for further exploration of convergence bounds.

PREREQUISITES
  • Understanding of parametric series and convergence criteria
  • Familiarity with the root and ratio tests for series convergence
  • Knowledge of logarithmic and exponential functions
  • Basic calculus concepts related to limits and sequences
NEXT STEPS
  • Study the application of the root test in series convergence
  • Investigate the ratio test and its limitations in parametric series
  • Explore advanced convergence criteria such as the integral test
  • Learn about the behavior of sequences involving logarithmic transformations
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Mathematicians, students studying advanced calculus, and anyone interested in the convergence of series and sequences.

Fochina
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Homework Statement
find for what ## \alpha ## the series converges
Relevant Equations
$$\sum_{n}\left ( \sqrt[n]{n}-\sqrt[n]{2} \right )^\alpha $$
It is clear that the terms of the sequence tend to zero when n tends to infinity (for some α) but I cannot find a method that allows me to understand for which of them the sum converges. Neither the root criterion nor that of the relationship seem to work. I tried to replace ##\sqrt[n]{n}## with ##e^{\frac{ln n}{n}}## but then I don't understand how to proceed.
 
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The result seems to be ##\alpha \geq 1##. So you could work with a convergent upper and a divergent lower bound which gives you some flexibility to change the function.
 

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