# Prove that $\sum\limits_{k=1}^{\infty} a_k$ is convergent if $\lim\limits_{k\to\infty} k(1 - \frac{a_{k+1}}{a_k})= \alpha$, where $\alpha > 1$

Prove that $\sum\limits_{k=1}^{\infty} a_k$ is convergent if $\lim\limits_{k\to\infty} k(1 - \frac{a_{k+1}}{a_k})= \alpha$, where $\alpha > 1$ and $k = 1, 2, ...$

This is my progress so far:

$k(1-\frac{a_{k+1}}{a_k}) = \alpha$

$\iff a_{k+1} = (1-\frac{a}{k})a_k$

By this we can see that $a_k$ is decreasing.

• This is Raabe's test. – Lord Shark the Unknown May 27 '17 at 11:23
• Pick $b$ in $(1,a)$ and show recursively that there exists $c$ and $n_0$ such that, for every $n\geqslant n_0$, $$a_n\leqslant cn^{-b}$$ To do that, it will help you to show that, for every $k$ large enough, $$\left(1-\frac{b}k\right)\frac1{k^b}\leqslant\frac1{(k+1)^b}$$ – Did May 27 '17 at 11:31