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Use the Root Test to determine whether the series is convergent or divergent.

$ \displaystyle \sum_{n = 2}^{\infty} \frac {( - 1)^{n-1}}{( \ln n)^n} $

Convergent

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Missouri State University

Baylor University

University of Michigan - Ann Arbor

Idaho State University

let's use the root test to determine whether or not the Siri's convergence. So let's call this our Anne then the root tests requires that we look at the limit as n goes to infinity and to root absolute value. And so, in our case, let me go ahead and use this fact from algebra that you could always write. The end through is just the rational exponents one over it. So that's what I'LL do here with this and through. So let's write the limit and goes to infinity So already I'll just write that is the one over any year Oops! Sorry about that. Using this fact over here and then absolute value of And so go ahead and take the absolute value of this fraction. The numerator just becomes a one the denominator natural log of end to the end power. And here I don't need absolute value anymore in the denominator, because natural log of end, it's bigger than zero. If Anne is bigger than or equal to two. And that's exactly the case that we're in for this problem. So now I could we could think of the inside here. This could be really in as well. One is equal to one to the end power. And we could write one over once in the end over natural life and to the end as one over natural log of end all to the end power. Then we're still raising that to the one over. And and now we use the fax from algebra that if you haven't exponents eight of the bee and if you raise that to another exponents E, then that's the same is just raising a to the B times C. So here this will be my baby and my cease all just multiplied those together. And when we do that, they just cancel auto one and we're left over with one over natural log. Since natural law goes to infinity, the fraction goes to zero. That's less than one. So we conclude that the Siri's convergence by the root test