## Calculus 8th Edition

Using the test for divergence that is stated in page 753 of the textbook we will take the limit of the corresponding sequence to the given series. Meaning, we will solve: $\lim\limits_{n \to \infty} \frac{k^{2}}{k^{2}-2k+5}$, to solve this we take the highest power of $k$ from the numerator and the denominator. In this case, we factor out $k^{2}$ and the result looks like: $\lim\limits_{n \to \infty} \frac{1}{1-\frac{2}{k}-\frac{5}{k^{2}} }$ and taking $k$ towards $\infty$, the fractions with $k$ at their denominator become zero and what remains is: $1$ $\ne$ 0 . Thus by test for divergence our series diverges.