Convergence Criteria for Series with Exponential and Polynomial Terms

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SUMMARY

The series \(\sum_{n=1}^{\infty} \frac{1}{e^n n^x}\) converges for all real values of \(x\). The user applied the ratio test, calculating \(\frac{U(n+1)}{U_n} = \frac{n^x}{e (n+1)^x}\) and derived the condition \(x > \frac{1}{\ln(n/(n+1))}\). However, as \(n\) approaches infinity, \(\ln(n/(n+1))\) approaches zero, leading to the conclusion that the series converges for any \(x\). The discussion emphasizes careful algebraic manipulation when applying convergence tests.

PREREQUISITES
  • Understanding of series convergence tests, specifically the ratio test.
  • Familiarity with exponential functions and logarithmic properties.
  • Basic knowledge of limits and their behavior as \(n\) approaches infinity.
  • Proficiency in manipulating inequalities and algebraic expressions.
NEXT STEPS
  • Study the Ratio Test for series convergence in more detail.
  • Explore the properties of logarithms, particularly in limits.
  • Learn about other convergence tests, such as the Root Test and Comparison Test.
  • Investigate the behavior of exponential functions in series and their convergence criteria.
USEFUL FOR

Mathematics students, educators, and anyone studying series convergence, particularly those working with exponential and polynomial terms in calculus.

wanchosen
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I have been stuck on this question for a while and have not got much success from class mates...can somebody please help?

Using an appropriate convergence test, find the values of x \in R for which the following series is convergent:

\sumn=1n \frac{1}{e^n * n^x}

So,

Un = \frac{1}{e^n * n^x}

and

U(n+1) = \frac{1}{e^(n+1) * (n+1)^x}

then

U(n+1)/Un = \frac{n^x}{e * (n+1)^x}

If the series is convergent:

\frac{n^x}{(n+1)^x} < e

ln(\frac{n^x}{(n+1)^x}) < 1

x * ln(\frac{n}{n+1}) < 1

n/n+1 is always positive but less than one, so the Ln term is negative,

So

x > \frac{1}{ln(n/n+1)}

But the limit of this as n tends to infininty, is infinity, which doesn't make sense. Can I solve this for x, independent of n?
 
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Your trouble is in the last step. You divide both sides by the log without realizing that as n->+inf, it is equal to zero.

Clearly, and you did all the work mind you, this is true for any x at all.

Reason? |x ln(n/n+1)| < 1 for all x you can think of... since any real number times zero equals zero, and this is less than 1 in absolute value.

Good job! Careful about that algebra, though...
 
Thanks!
 

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