Interval of Convergence and radicals

Click For Summary
SUMMARY

The discussion focuses on finding the interval of convergence for the series \(\sum _{n=1}^{\infty } \frac{(-1)^n (x+2)^n}{3^n\sqrt{n}}\). The limit of the ratio test is determined to be \(\frac{(x+2)}{3}\) as \(n\) approaches infinity. The participants confirm that \(\lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+1}} = 1\), which is crucial for simplifying the limit calculation. This establishes the necessary steps to analyze the convergence of the series.

PREREQUISITES
  • Understanding of series convergence tests, specifically the Ratio Test.
  • Familiarity with limits and their properties in calculus.
  • Knowledge of radical expressions and their simplifications.
  • Basic proficiency in using mathematical software like Mathematica for limit calculations.
NEXT STEPS
  • Study the Ratio Test in detail to apply it to various series.
  • Learn about the concept of interval of convergence for power series.
  • Explore the use of Mathematica for solving calculus problems, particularly limits.
  • Investigate the properties of limits involving radicals and their applications in convergence tests.
USEFUL FOR

Students studying calculus, particularly those focusing on series and convergence, as well as educators seeking to clarify concepts related to power series and limit calculations.

justtip
Messages
1
Reaction score
0

Homework Statement



Find the interval of convergence:

[tex]\sum _{n=1}^{\infty } \frac{(-1)^n (x+2)^n}{3^n\sqrt{n}}[/tex]


Homework Equations





The Attempt at a Solution


[tex]\lim_{n\to \infty } |\frac{(x+2)^{n+1}}{3^{n+1}\sqrt{n+1}}*\frac{3^n\sqrt{n}}{(x+2)^n}| = \lim_{n\to \infty } |\frac{(x+2)\sqrt{n}}{3\sqrt{n+1}}|[/tex]


This is where I'm stuck. Mathematica says the limit is (x+2)/3, but I'm not sure how to get there. Only thing I can think of is if infinity is substituted for n, the square root of infinity cancels out? I can work the rest out myself, just need to know how to get to (x+2)/3.

Thanks.
 
Physics news on Phys.org
welcome to pf!

hi justtip! welcome to pf! :smile:
justtip said:
Mathematica says the limit is (x+2)/3, but I'm not sure how to get there.

you're trying to prove that lim √n/√(n+1) = 1

that's the same as lim √(n/(n+1)) = 1,

which is the same as lim n/(n+1) = 1 :wink:
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
17
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K