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

[AceK]

Homework Statement

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

Homework Equations

a_{n} converges if \stackrel{lim}{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1.

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 \stackrel{lim}{n\rightarrow\infty}\left|\frac{2^{n}-n}{2^{n+1}-n-1}\right|. Is it possible to just say this converges, or is there a mathematical trick to simplifying the ratio? Thanks in advance for the help!

 

[ystael]

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.

 

[Mark44]

Or you could continue with what you've done using the Ratio test. Factor 2ⁿ out of each term in the numerator and denominator.

 

[spyrustheviru]

divide the numerator and denominator with 2^n and the result will be easy to calculate

 

[AceK]

ystael: I thought long and hard about the comparison test, but the only thing I could think of to use as a comparison is \frac{1}{2^{n}}, 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 2^{n} results in \lim_{n\rightarrow\infty}\frac{1-n2^{-n}}{2-(n+1)2^{-n}}, correct? I'm not sure I understand how this makes the problem any easier.

 

[ystael]

ystael: I thought long and hard about the comparison test, but the only thing I could think of to use as a comparison is \frac{1}{2^{n}}, 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 \sum \frac1{2^n} a little bit to find a series that is greater than the original series, and converges for the same reason that \sum \frac1{2^n} converges.