Question #b6d9b

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Answer 1 · 2017-02-05T18:02:43

The sequences are convergent to #0# and the series are divergent.

Considering the series:

For #a,b in RR^+# we have for #n > 1#

#1/(max(a,b)(n-1)) < 1/(an-b)# (supposing that #an-b ne 0#) and we know that

#sum_(n=2)^oo1/n# diverges

So for #n^@ > max(1,floor(b/a))# we have

#1/(max(a,b)) sum_(n = n^@)^oo 1/(n-1) < sum_(n = n^@)^oo1/(an-b)#

but

#sum_(n = n^@)^oo 1/(n-1)# diverges so #sum_(n = n^@)^oo1/(an-b)#

diverges also.