I was doing some exercises on the nature of infinite series when I came across this one intriguing series - ∞∑i=1(−1)i−1ln(i+1). I tried to solve it with D'Alembert's ratio test and came at the solution that this series is convergent as lim comes out to be \lim\limits_{n \to \infty}|\frac{-\ln(n+1)}{\ln(n+2)}| = \lim\limits_{n \to \infty}\frac{\ln(n+1)}{\ln(n+2)} which is smaller than 1 as \ln x is strictly increasing. So, the - sign doesn't really make any difference. But the answer in the textbook is that this series is conditionally convergent, meaning that it wouldn't have been convergent if not for the - sign. Can anyone explain this to me?
Subscribe to:
Post Comments (Atom)
real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}
How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
Ok, according to some notes I have, the following is true for a random variable X that can only take on positive values, i.e P(X \int_0^...
-
Self-studying some properties of the exponential-function I came to the question of ways to assign a value to the divergent sum $$s=\sum_{k=...
-
I use Euclidean Algorithm: 4620 = 101 * 45 + 75. long story short. I get 3 = 2 * 1 + 1. After that 2 = 1 * 2 + 0. gcd(101,4620) = 1. So I us...
No comments:
Post a Comment