With the power series Sigma ln n over n all multiplied by x to the nth: show interval of convergence is [-1,1)

Answer 1

an = x^n* ln n
By the ratio test applied to Ia_nI
I a_n+a/a_nI= IxI ln(n+1)/ln(n) with limit IxI
so if
IxI<1 the seies is absolutely convergent
-1 At the extremes lim a_n is NOT 0 so it doe´s diverge at x=+-
and outside (-1,1)

Answer 2

Use the ratio try. check out lim n->? |((ln (n + a million) * x^(n + a million)) / (n + a million)) / ((ln (n) * x^n) / n)|. Rewrite the quotient of fractions as a product: = lim n->? |((ln (n + a million) * x^(n + a million)) / (n + a million)) * (n / (ln (n) * x^n))|. Cancel x^n: = lim n->? |((ln (n + a million) * x) / (n + a million)) / (n / ln (n))|. Pull each and every thing with an 'n' in it outdoors actual the cost: = lim n->? (ln (n + a million) * n)/(ln (n) * (n +a million)) * |x| element: = lim n->? (ln (n + a million) / ln (n)) * (n / (n + a million)) * |x| Now, of course n / (n + a million) has a tendency to a million. It seems that ln (n + a million) / ln (n) additionally has a tendency to a million (you an teach this using L'well being facility's Rule, as an occasion). considering that those 2 factors tend to a million, then we get = lim n->? a million * a million * |x| = |x|. simply by fact the ratio try utilized to the series supplies |x|, then the series converges whilst |x| < a million and diverges whilst |x| > a million. So the radius of convergence is a million.

Answer 3

use the ratio test.
limit n-> infinity = abs (f(n+1)/f(n)) < 1; solve for x