Root Test: Convergence of Series

In summary, using the comparison test, it can be concluded that the series \sum_{n=2}^{\infty}\frac{1}{(\ln n)^{\ln (\ln n)}}=\sum_{n=2}^{\infty}\frac{1}{e^{\ln (\ln n)\ln (\ln n)}} diverges. This is because \ln^2(\ln n)<\ln n<e^{\ln^2(\ln n)}, which leads to \frac{1}{e^{\ln^2(\ln n)}}>\frac{1}{n}, indicating that the series is larger than the divergent harmonic series.
  • #1
azatkgz
186
0

Homework Statement


Determine whether the series converges or diverges.

[tex]\sum_{n=2}^{\infty}\frac{1}{(\ln n)^{\ln (\ln n)}}=\sum_{n=2}^{\infty}\frac{1}{e^{\ln (\ln n)\ln (\ln n)}}[/tex]





The Attempt at a Solution



for [tex]u=\ln (\ln n)[/tex]

[tex]\sum_{u=\ln (\ln 3)}^{\infty}\frac{1}{e^{u^2}}[/tex]

from Root Test

[tex]\lim_{u\rightarrow\infty}\sqrt{\frac{1}{e^{u^2}}}=\lim_{u\rightarrow\infty}\frac{1}{e^u}=0<1[/tex]
series converges
 
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  • #2
That u substitution does not give back the original series. ln(ln(3))+1 does noe equal ln(ln(4)).

Try simply the comparison criterion. Hint: (logx)²/x -->0, so (logx)²<x for x large enough.
 
  • #3
quasar987 said:
Try simply the comparison criterion. Hint: (logx)²/x -->0, so (logx)²<x for x large enough.

[tex]\ln^2x<x[/tex]

for x=lnn

[tex]\ln^2(\ln n)<\ln n[/tex]

[tex]e^{\ln^2(\ln n)}<e^{\ln n}=n[/tex]

[tex]e^{\ln^2(\ln n)}<n\rightarrow \frac{1}{e^{\ln^2(\ln n)}}>\frac{1}{n}[/tex]

diverges
 
  • #4
good!
 

1. What is the root test for convergence of series?

The root test is a method used to determine the convergence or divergence of a series. It involves taking the nth root of the terms in a series and evaluating the limit as n approaches infinity. If the limit is less than 1, the series is convergent. If the limit is greater than 1, the series is divergent. If the limit is equal to 1, the test is inconclusive and another method must be used.

2. How is the root test different from the ratio test?

Both the root test and ratio test are used to determine the convergence or divergence of a series. However, the root test involves taking the nth root of the terms in a series while the ratio test involves taking the limit of the absolute value of the ratio of successive terms. The root test is generally easier to apply and often gives a quicker result, but the ratio test can be used in cases where the root test does not yield a conclusive result.

3. Can the root test be used for both infinite and finite series?

Yes, the root test can be used for both infinite and finite series. However, it is more commonly used for infinite series as it is designed to determine the behavior of a series as the number of terms approaches infinity.

4. When should the root test be used instead of the comparison test?

The root test is generally used when the series involves exponential or factorial terms, as it is specifically designed to handle these types of series. The comparison test is better suited for series with simpler terms, such as polynomials. Additionally, the root test can be used when the comparison test yields an inconclusive result.

5. Are there any limitations to the root test?

Yes, there are some limitations to the root test. One limitation is that it can only be used for series with positive terms. Additionally, it may not always yield a conclusive result, in which case another method must be used. It is also important to note that the root test is a sufficient but not necessary condition for convergence, meaning that a series may pass the root test but still converge by another method.

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