MHB Proving Series Convergence: Comparing $\sum y_n$ with $\sum \frac{y_n}{1+y_n}$

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SUMMARY

The discussion focuses on proving the convergence of the series $\sum_{n=1}^{\infty} y_n$ given that the series $\sum_{n=1}^{\infty} \frac{y_n}{1+y_n}$ converges. It is established that since $y_n \geq \frac{y_n}{1+y_n}$, the convergence of the latter implies the convergence of the former. The comparison test is suggested as a method for proving this relationship, leveraging the fact that as $n$ approaches infinity, $\frac{y_n}{1+y_n}$ approaches zero, which indicates that $y_n$ also approaches zero.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with the comparison test in series analysis
  • Knowledge of limits and bounding techniques in sequences
  • Basic concepts of sequences and their properties
NEXT STEPS
  • Study the comparison test for series convergence in detail
  • Explore the implications of the limit $\frac{y_n}{1+y_n} \to 0$ on the sequence $(y_n)$
  • Investigate bounding techniques for sequences and their applications in convergence proofs
  • Review examples of series that utilize similar convergence arguments
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Mathematicians, students studying real analysis, and anyone interested in series convergence proofs and comparison techniques.

evinda
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Hello! (Wave)
We have a sequence $(y_n)$ with $y_n \geq 0$.
We assume that the series $\sum_{n=1}^{\infty} \frac{y_n}{1+y_n}$ converges. How can we show that the series $\sum_{n=1}^{\infty} y_n$ converges?

It holds that $y_n \geq \frac{y_n}{1+y_n}$.

If we would have to prove the converse we could use the comparison test. Could you give me a hint what we can do in this case?
 
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From $\dfrac{y_n}{1+y_n}\to0$ we can conclude that $y_n\to0$. Therefore, $1+y_n$ can be eventually bounded from above, say, by $3/2$. So, $\dfrac{y_n}{1+y_n}$ can be bounded from below by $\dfrac23y_n$.
 
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