I'm doing Problem II.7.5 in textbook Analysis I by Amann/Escher.

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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 $0$.

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

1 Answer 1


Your proof looks good, well done!


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