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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.

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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 $\sum_{k\geq 2} a_k$.

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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.

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