# Proving convergence of a sum give an inequality

Let $$\alpha>0$$ be a real number and let $$N$$ be an integer. Let's assume that $$a_n$$ is a positive sequence such that for every $$n>N$$: $$(n-1)a_n-na_{n+1}\ge\alpha{a_n}$$

I want to prove that the sequence of partial sums $$s_k=\sum_{n=1}^k{\alpha a_n}$$ is bounded and from that to prove that $$\sum_{n=1}^\infty{a_n}$$ converges.

I've managed to prove that $$a_n$$ is monotone decreasing:

$$(n-1)a_n-na_{n+1}\ge\alpha{a_n}$$

$$(n-1)a_n-\alpha a_n\ge na_{n+1}$$

$$na_n>a_n(n-1-\alpha)\ge na_{n+1}$$

$$a_n>a_{n+1}$$

I thought of using the following lemma:

If $$(na_n)_{n=1}^\infty$$ is a monotone increasing sequence then $$\sum_{n=1}^\infty{a_n}$$ is divergant.

Since you wrote $na_n>a_n(n-1-\alpha)$ I'm assuming the $a_i$ are $\geq 0$
Summing the inequalities $(n-1)a_n-na_{n+1}\ge\alpha{a_n}$ from $n=N+1$ to $n=M$ yields $$Na_{N+1}\geq \alpha \sum_{k=N+1}^Ma_k +Ma_{M+1}\geq \alpha \sum_{k=N+1}^Ma_k$$
Since $Na_{N+1}\geq \alpha \sum_{k=N+1}^Ma_k$ holds for every $M\geq N+1$, $\sum_{k\geq N} a_k$ converges, and so does $\sum_{k\geq 2} a_k$.
We have: $$a_{n+1}\leq \frac{n-1-\alpha}{n}\,a_n$$ hence, by assuming $n\geq \alpha+1$ and $a_n\geq 0$, $$a_{n+1} \leq \exp\left(-\frac{\alpha+1}{n}\right)\, a_n$$ so, by induction, $$a_{n+K} \leq a_n\cdot\exp\left(-(\alpha+1)\sum_{m=n}^{n+K-1}\frac{1}{m}\right)\approx a_n\cdot \left(\frac{n}{n+K}\right)^{\alpha+1}$$ but the series $\sum_{K\geq 1}\frac{1}{K^{\alpha+1}}$, given $\alpha>0$, is convergent by the p-test, so $\sum_{m\geq 1} a_m$ is convergent, too, provided that $a_{\left\lceil \alpha+1\right\rceil}\geq 0$. Have a look at Raabe's test.