Let $(x_n)\downarrow 0$ and $\sum x_n\to s$. Then $(n\cdot x_n)\to 0$

Let $$(x_n)$$ a decreasing sequence and $$\sum x_n\to s$$. Then $$(n\cdot x_n)\to 0$$

Check my proof please, Im not completely sure about it correctness.

If $$\sum_{k=h}^\infty x_k= s$$ then we can rewrite the sum for starting index $$1$$ with the change $$k-h=j$$, then

$$\left(\sum_{j=1}^n x_j\right)-n\cdot x_n=\sum_{j=1}^n (x_j-x_n)$$

Then taking limits

$$\color{red}{\lim_{n\to\infty}\left[\left(\sum_{j=1}^n x_j\right)-n\cdot x_n\right]}=\lim_{n\to\infty}\sum_{j=1}^n (x_j-x_n)=\sum_{j=1}^\infty (x_j-0)=\color{red}{\lim_{n\to\infty}\sum_{j=1}^n x_j}=s$$

where I used the fact that $$(x_n)\to 0$$. Thus equating the colored expressions this implies that $$\lim_{n\to\infty} nx_n=0$$.

The proof, to my eyes, seems correct but I dont needed in any moment to use the fact that $$(x_n)$$ is a monotonic sequence so it is possible that I make a mistake somewhere or that the proof is incorrect.

My second attempt

Because $$\sum x_k$$ converges and is positive (cause $$(x_n)\downarrow 0$$) we can write

$$\sum_{k=n+1}^{2n+m}x_k=\left|\sum_{k=n+1}^{2n+m}x_k\right|<\epsilon/2,\quad \forall n,m\ge N$$

Then, cause $$(x_n)$$ is decreasing

$$(n+m)x_{2n+m}\le\sum_{k=n+1}^{2n+m}x_k<\epsilon/2\\\implies (2n+m)x_{2n+m}\le2(n+m)x_{2n+m}<\epsilon,\quad\forall n,m\ge N$$

Because $$m$$ is arbitrary setting $$M=2N>N$$ we can finally write

$$nx_n<\epsilon,\quad\forall n\ge M$$

It is this proof correct? Thank you.

• The step $$\lim_{n \to \infty}\sum_{j=1}^n (x_j - x_n) = \lim_{n \to \infty}\sum_{j=1}^n \lim_{n \to \infty}(x_j - x_n)$$ Seems a bit suspect, but I can't say with certainty that it's wrong – Omnomnomnom Oct 25 '16 at 0:07
• If $x_n=\frac{(-1)^n}n$, $\sum_nx_n$ converges, and $x_n\to 0$ (though not monotonically), but $nx_n=(-1)^n$ does not converge to $0$. This shows that your argument can’t be right, since nothing in it rules out this example. – Brian M. Scott Oct 25 '16 at 0:17
• @Omnomnomnom. It IS a wrong step. The attempted proof does not use the monotonicity of $x_n$. But the result does not hold for all non-monotonic $x_n$. E.g. if $x_{(n^4)}=1/n^2$, and $x_n=0$ when $n$ is not a $4$th power. – DanielWainfleet Oct 25 '16 at 0:22
• Ok, it is clear that this proof is not correct. I will investigate a different proof, this time totally formal. Thank you to all. – Masacroso Oct 25 '16 at 0:23
• On top of my head, I think your proof can be adjusted with Beppo-Levi theorem. However, it is not a result which is taught in the first course of analysis. – user228113 Oct 25 '16 at 0:25

If $\sum_{k=1}^\infty x_k = s,$ then by the Cauchy criterion for any $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that if $n > N$ we have $\sum_{k= n+1}^{2n} x_k < \epsilon.$ Since $(x_n)$ is decreasing and positive,$\sum_{k= n+1}^{2n} x_k > nx_{2n}.$
• @Masacroso: You might add -- with $m=0$ we see $2nx_{2n} \to 0$ and with $m = 1$ we see $(2n+1)x_{2n+1} \to 0$. Since both subsequences converge it follows that $nx_n \to 0$. Well done. – RRL Oct 25 '16 at 14:22
Here's a correct proof: apply summation by parts to $\sum a_n b_n$, where $b_n = 1$ for all $n$.