Exists some #x# for which #sum_{k=0}^oo [log(-x)]^-k# converges, and what is the sum?

Question

450 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-16T12:41:01

#x < -e#

Assuming #log y = log_e y# and making #y = -x > 0# we have

#S(y)=sum_{k=0}^oo [log(y)]^-k = sum_{k=0}^oo1/log^k(y)#

Now there are infinite #y > 0# such that #S(y)# converges.

For instance, with #log(y) = lambda# we have

#S(e^lambda) = sum_{k=0}^oo1/lambda^k= lim_(n->oo)((1/lambda^(n+1)-1)/(1/lambda-1))# which converges to

#1/(1-1/lambda)= lambda/(lambda-1)#

for #lambda > 1#

or #y = e^lambda rArr y > e rArr -x > e rArr x < -e#