Like

Report

Find the radius of convergence and interval of convergence of the series.

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

$$I=(-\infty, \infty)$$

You must be signed in to discuss.

Missouri State University

Oregon State University

Idaho State University

to figure out the radius of convergence will first just look at where the Siri's converges So we'LL use the ratio. Tests take limit as n goes to infinity of absolute value of A and plus one over a n where a. And is this whole thing here, including the X values. So that's limit as n goes to infinity of absolute value of X minus two to the end plus one divided by in plus one the power of in plus one divided by a n, which is something as multiplying by the reciprocal. Okay, so x minus two to the impulse one divided by X minus two to the end. That just simplifies tow X minus two. Okay, that's one of the nice thing about having a common base. And here we don't quite have a common exponents, but in plus one of the power of in plus one can be written as in plus one to the N times in plus one. So if we do that, then we can rewrite this thing like this and now limit as n goes to infinity of in over in plus one to the end. This is one over e So now we have limit as n goes to infinity of absolute value of X minus two over in plus one. And this value is one over e. So he is just some number here and one over in plus one goes to infinity. Doesn't matter what exes as long as exit any finite number. This is just going to go to zero, which is certainly less than one. We're going to get convergence regardless of what value of X. We're looking at here. So the radius of convergence is infinity in the interval of convergence is minus infinity to infinity.