I need help solving this exercise about infinite series from Robert Bartle's "Elements of real analysis" book.
Let $x_{n}>0$ for $n \in \mathbb{N}$ and suppose that $n\left( 1-\frac{x_{n+1}}{x_{n}} \right)=a+\frac{k_{n}}{n^{p}}$ where $p>0$ and $k_{n}$ is bounded. Then the series $ \sum_n x_{n}$ converges if $a> 1$ and diverges if $ a<1 $.
One has $$ \frac{x_{n+1}}{x_n} = 1 - \frac{a}{n} + \frac{k_n}{n^{p+1}} = 1 - \frac{a}{n} + O \left( \frac{1}{n^{p+1}}\right)$$
Let's define $v_n = n^a x_n$. One has $$\frac{v_{n+1}}{v_n} = \left( 1 + \frac{1}{n}\right)^a \frac{x_{n+1}}{x_n} = \left( 1 + \frac{1}{n}\right)^a \left( 1 - \frac{a}{n} + O \left( \frac{1}{n^{p+1}}\right)\right)$$ $$=\left( 1 + \frac{a}{n} + O \left( \frac{1}{n^2}\right)\right) \left( 1 - \frac{a}{n} + O \left( \frac{1}{n^{p+1}}\right)\right) =1 + O \left( \frac{1}{n^{\gamma}}\right)$$
where $\gamma = \min \left( 2, p+1\right) > 1$. So $$\ln \left( \frac{v_{n+1}}{v_n}\right) = O \left( \frac{1}{n^{\gamma}}\right)$$
So the series $\ln \left( \frac{v_{n+1}}{v_n}\right)$ converges, so the sequence $(\ln (v_n))$ converges, so the sequence $(v_n)$ converges : denoting $K = \lim v_n$, you get directly $$u_n \sim \frac{K}{n^a}$$
which gives you the desired results.