MHB  Use the Ratio Test for Convergence/Divergence

AI Thread Summary
The discussion focuses on applying the ratio test to determine the convergence or divergence of the series defined by the expression (1 - (1/k))^(3k) as k approaches infinity. The limit of the ratio of consecutive terms is calculated, yielding a result of 1, which is inconclusive for the ratio test. However, further analysis shows that the limit of the series terms approaches e^(-3), which is not zero. Therefore, the series is concluded to be divergent based on the term test. The application of both the ratio test and the term test illustrates the importance of multiple methods in assessing series convergence.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
I quote a question from Yahoo! Answers

Use the ratio test to solve:? 1) k=1--> inf (1-(1/k))^(3k) 2)Is it convergent or divergent?

I have given a link to the topic there so the OP can see my response.
 
Mathematics news on Phys.org
We have

$\displaystyle\lim_{k\to +\infty}\frac{u_{k+1}}{u_k}=\lim_{k\to +\infty}\frac{\left(1-\frac{1}{k+1}\right)^{3k+3}}{\left(1-\frac{1}{k}\right)^{3k}}=\lim_{k\to +\infty}\left(1-\frac{1}{k+1}\right)^3\cdot\lim_{k\to +\infty}\frac{\left(1-\frac{1}{k+1}\right)^{3k}}{\left(1-\frac{1}{k}\right)^{3k}}$

$=1\cdot\displaystyle\lim_{k\to +\infty}\left(\frac{k^2}{k^2-1}\right)^{3k}=\ldots=1\cdot 1=1\text{ (inconclusive)}$

But $\displaystyle\lim_{k\to +\infty}u_k=\displaystyle\lim_{k\to +\infty}\left(1-\frac{1}{k}\right)^{3k}=\ldots=e^{-3}\ne 0$ so, (term test) the series is divergent.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Back
Top