Help on comparison test / determining Bn

Question

This question has me stumped. I was thinking of using the comparison test, then using the integral test on bn. But I can't even figure out what to make bn in this particular situation as the form of this series doesn't appear familiar to me.

Answer 1

You can use the root test$$\lim_{k\to\infty}\sqrt[k]{\frac k{10^k-k^2}}=\lim_{k\to\infty}\frac{\sqrt[k]k}{10\sqrt[k]{1-\frac{k^2}{10^k}}}=\frac1{10}<1.$$