Convergence testing of the series 1/(2^n-n)

In summary, Homework Equations converges if \stackrel{lim}{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1.
  • #1
AceK
6
0

Homework Statement


Determine whether [tex]\sum\frac{1}{(2^{n}-n)}[/tex] from n=1 to infinity converges or diverges.


Homework Equations


a[tex]_{n}[/tex] converges if [tex]\stackrel{lim}{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1[/tex].


The Attempt at a Solution


I'm really having a tough time knowing where to start on this one. I've tried the integral test, the comparison test, and the ratio test with no luck. Wolfram Alpha says it converges by the ratio test, but I know of no way to simplify [tex]\stackrel{lim}{n\rightarrow\infty}\left|\frac{2^{n}-n}{2^{n+1}-n-1}\right|[/tex]. Is it possible to just say this converges, or is there a mathematical trick to simplifying the ratio? Thanks in advance for the help!
 
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  • #2
Think about the relative strength of the two terms in the denominator, and try to find a crude estimate to use the comparison test on.
 
  • #3
Or you could continue with what you've done using the Ratio test. Factor 2n out of each term in the numerator and denominator.
 
  • #4
divide the numerator and denominator with 2^n and the result will be easy to calculate
 
  • #5
ystael: I thought long and hard about the comparison test, but the only thing I could think of to use as a comparison is [tex]\frac{1}{2^{n}}[/tex], which is greater than the original function, and it converges. i also tried variants of replacing 2 with other bases, and I found none that were either convergent and greater than f(x) or divergent and less than f(x).

Mark44: Factoring out [tex]2^{n}[/tex] results in [tex]\lim_{n\rightarrow\infty}\frac{1-n2^{-n}}{2-(n+1)2^{-n}}[/tex], correct? I'm not sure I understand how this makes the problem any easier.
 
  • #6
AceK said:
ystael: I thought long and hard about the comparison test, but the only thing I could think of to use as a comparison is [tex]\frac{1}{2^{n}}[/tex], which is greater than the original function, and it converges. i also tried variants of replacing 2 with other bases, and I found none that were either convergent and greater than f(x) or divergent and less than f(x).

I think you mean "less" rather than "greater". But you can modify [tex]\sum \frac1{2^n}[/tex] a little bit to find a series that is greater than the original series, and converges for the same reason that [tex]\sum \frac1{2^n}[/tex] converges.
 

Q1: What is convergence testing?

Convergence testing is a process used to determine whether a series, or a sequence of numbers, will approach a finite limit or not.

Q2: Why is it important to test for convergence?

Testing for convergence helps us determine if a series is useful for calculations and if it will provide a meaningful result. It also helps us understand the behavior of the series and make predictions about its future behavior.

Q3: What is the series 1/(2^n-n)?

The series 1/(2^n-n) is a sequence of numbers where the denominator increases exponentially while the numerator remains constant. The series can also be written as (1/2)^n - 1/n.

Q4: How is the convergence of the series 1/(2^n-n) determined?

The convergence of this series can be determined by using various convergence tests such as the ratio test, root test, or comparison test. These tests compare the series to a known convergent or divergent series and help determine the behavior of the series in question.

Q5: What is the result of the convergence test for the series 1/(2^n-n)?

The series 1/(2^n-n) is a convergent series. This can be proven by using the ratio test or the comparison test, which both show that the series approaches a finite limit of 0. Therefore, the series 1/(2^n-n) is considered a useful and meaningful series for calculations.

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