Science:Math Exam Resources/Courses/MATH103/April 2012/Question 01 (a)/Solution 1

The difference between the highest exponents is 1.1-0.1 = 1. By the series p test (with p = 1) this series diverges.

More precisely we can prove

$\sum _{n=1}^{\mathrm{\infty }}\frac{n^{0.1}}{n^{0.99}+n^{1.1}+1}>\sum _{n=1}^{\mathrm{\infty }}\frac{n^{0.1}}{n^{1.1}+n^{1.1}+n^{1.1}}=\sum _{n=1}^{\mathrm{\infty }}\frac{n^{0.1}}{3n^{1.1}}=\frac{1}{3}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}=\mathrm{\infty }$

The last sum diverges because of the series p test with p=1.