Question

the series sum (-1)^n / ln(n+1) is divergent? conditionally convergent? absolutely convergent?

Answer 1

\[\sum_{n=1}^{\infty}\frac{ (-1)^n }{ \ln (n+1) }\]

Answer 2

this should be alternating type

Answer 3

isn't this absolutely convergent?

Answer 4

my note says ratio,root,absolute value, alternating series are absolutely convergent, but i don't know why..

Answer 5

the correct answer is conditionally convergent, if you take the absolute value of the series, you will get\[\sum_{n=1}^{\infty}\frac{1}{\ln(n+1)}\]try to test this series with the comparison test, compare with 1/n

Answer 6

incorrect you will compare this with 1/ n+1 and not 1/n

Answer 7

if you want to approximate , we can say 2.3 *(n+1) = ln (n+1), pull out 2.3 as constant and we are left with n+1

Answer 8

but the overall answer is correct

Answer 9

why is it incorrect? the inequality\[\ln (n + 1) < n\]is true for n > 0, which implies\[\frac{1}{\ln(n + 1)} > \frac{1}{n}\]for n > 0