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

convergence-divergenceproof-verificationreal-analysissequences-and-series

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 $0$.

Best Answer

Your proof looks good, well done!