Thursday, 18 August 2016

Ratio test for convergent series - Does that mean that the series 1+frac12+frac13+dotsb converges?






  1. An infinite series is convergent if from and after some fixed term, the ratio of each term to the preceding term is numerically less than some quantity which is itself numerically less than unity.
    Let the series beginning from the fixed term be denoted by
    u1+u2+u3+u4+,
    and let
    $$\frac{u_2}{u_1} where r<1.
    Then
    \begin{align*}
    &u_1+u_2+u_3+u_4+\dotsb\\

    &=u_1\left(1+\frac{u_2}{u_1}+\frac{u_3}{u_2}\cdot\frac{u_2}{u_1}+\frac{u_4}{u_3}\cdot\frac{u_3}{u_2}\cdot\frac{u_2}{u_1}+\dotsb\right)\\
    &\end{align*}
    that is, <u11r, since r<1.
    Hence the given series is convergent.




Does that mean that the series 1+12+13+ should be a convergent one?
Sn=11+12+13+14+
Here, we have

unun1=1/n1/(n1)=n1n=11nuun1<1Series should be convergent


Answer



You need unun1<r<1 for some r<1 and all n.



For any r<1 we can find r<unun1<1. So we can not find an appropriate r<1.




So we failed the hypothesis. The ratio test fails.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...