Question #b6d9b

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Answer 1 · 2017-02-05T18:02:43

The sequences are convergent to $0$ $0$ 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.