# Let $(x_{k})$ be a decreasing sequence such that $\sum x_{k}$ converges. Prove that $(k x_{k})$ is a null sequence

I'm doing Problem II.7.5 in textbook Analysis I by Amann/Escher. Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:

Since $$(x_k)$$ is decreasing, we have $$\sum_{k=N}^{n} x_k \ge (n-N+1) x_n$$ or equivalently $$\sum_{k=N}^{n} x_k + (N-1)x_n \ge n x_n$$ for all $$n >N$$.

By Cauchy's convergence test: given $$\epsilon >0$$ a, there exists $$N_1$$ such that $$\sum_{k=n}^m x_k < \epsilon/2$$ for all $$m>n \ge N_1$$. Since $$(x_k)$$ converges to $$0$$, there exists $$N_2$$ such that $$|x_k| < \epsilon/(2(N_1-1))$$ for all $$n \ge N_2$$. Take $$N = \max \{N_1,N_2\}$$, we have $$\sum_{k=N}^{n} x_k < \epsilon/2$$ and $$(N -1) |x_{n}| < \epsilon/2$$ for all $$n > N$$.

As such, $$n x_n \le \sum_{k=N}^{n} x_k + (N-1)x_n \le \epsilon/2 + \epsilon/2 = \epsilon$$ for all $$n > N$$. This implies that $$(k x_k)$$ converges to $$0$$.

• your proof looks good Aug 27, 2019 at 15:11
• Thank you so much @Riquelme! Could you please write your comment as an answer so that I can peacefully close this question? Aug 27, 2019 at 15:12