MHB Series Convergence with Comparison Test

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SUMMARY

The discussion focuses on the convergence of a series using the Comparison Test in Calculus III and Analysis. The key insight is that the first few terms of the series can be ignored, allowing analysis to begin with the term $\frac{1}{3+\sqrt{2}}$. The series terms are of the form $\frac{1}{3^n + \sqrt{n+1}}$, which can be compared to the geometric series $\left(\frac{1}{3}\right)^n$. Since $\frac{1}{3^n + \sqrt{n+1}} < \frac{1}{3^n}$ for all $n \geq 0$, the series converges by the Comparison Test.

PREREQUISITES
  • Understanding of series convergence and divergence
  • Familiarity with the Comparison Test in calculus
  • Knowledge of geometric series and their properties
  • Basic skills in manipulating algebraic expressions involving exponents and square roots
NEXT STEPS
  • Study the Comparison Test in detail, focusing on its applications in series convergence
  • Learn about geometric series and their convergence criteria
  • Explore advanced topics in series analysis, such as the Ratio Test and Root Test
  • Practice problems involving series convergence with varying types of terms
USEFUL FOR

Students in Calculus III and Analysis, educators teaching series convergence, and anyone seeking to deepen their understanding of series and convergence tests in mathematics.

Zoey93
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Hey,

I am working on Calculus III and Analysis, I really need help with this one problem. I am not even sure where to begin with this problem. I have attached my assignment to this thread and the problem I need help with is A. Thank you!
 

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Zoey93 said:
Hey,

I am working on Calculus III and Analysis, I really need help with this one problem. I am not even sure where to begin with this problem. I have attached my assignment to this thread and the problem I need help with is A. Thank you!
Here are a couple of hints to get you started.

1. The convergence or divergence of a series is not affected by the first few terms. So you can neglect the first three terms of the series, and concentrate on those starting with the term $\frac1{3+\sqrt2}.$

2. From that point onwards, the denominator of each term consists of two elements: a power of 3 and a square root. One of those elements is going to be much larger than the other. So think about what the series would look like if you neglected the smaller element.
 
Each of the important terms are of the form $\displaystyle \begin{align*} \frac{1}{ 3^n + \sqrt{n+1}} \end{align*}$, and since $\displaystyle \begin{align*} 3^n + \sqrt{n + 1} > 3^n \end{align*}$ for $\displaystyle \begin{align*} n \geq 0 \end{align*}$, that means $\displaystyle \begin{align*} \frac{1}{3^n + \sqrt{n + 1}} < \frac{1}{3^n} = \left( \frac{1}{3} \right) ^n \end{align*}$. So you can make a comparison to a simple geometric series :)
 
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