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I need a little hint in a proof on convergence of a sequence. The problem I have is: Suppose $(x_n)$ is a monotone decreasing sequence of real numbers such that $\sum x_n$ converges, prove that:

$$\lim_{n \to \infty} nx_n=0$$

My idea is: proving this fact aims to prove that given $\epsilon >0$ there's some natural $n_0 \in \mathbb{N}$ such that if $n >n_0$ we have $|nx_n|<\epsilon$.

Now, I know that $\sum x_n$ converges, so that given $\epsilon'>0$ there's some $k_0 \in \mathbb{N}$ such that if $k > k_0$ we have:

$$\left|\sum_{i=1}^{k}x_i - S\right|<\epsilon'$$

Where $S = \sum x_n$. Now, since $(x_n)$ is monotone decreasing we know that we must have $x_1 > \cdots > x_k$ so that the sum of all $x_i$ should be less or equal to $k x_k$. So I know that I have:

$$\left|\sum_{i=1}^{k}x_i - S\right|\leq\left|\sum_{i=1}^{k}x_i\right|+|S|\leq|kx_k|+|S|$$

I feel that the proof will come from this, however I'm stuck at this point. Can someone give just a little hint on how to proceed from here?

Thanks very much in advance for your help.

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  • $\begingroup$ I have no idea if it's a good answer or not but my first thought went to "compare it to $\frac{1}{n}$"... Again, I have no idea of it could be possible and correct as it is not general. $\endgroup$ – mwoua May 6 '13 at 20:46
  • $\begingroup$ If you can prove the limit exists, then it must be zero by the limit comparison test with $\frac{1}{n}$. $\endgroup$ – vadim123 May 6 '13 at 20:47
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    $\begingroup$ Another strategy: It's a general fact that for all nonnegative sequences $(a_n)$ if $\sum_{n=1}^\infty a_n < \infty$, then $\liminf\limits_{n\to\infty} na_n = 0$ (hint: the contrapositive is easy to prove). Then you only need to show that $\limsup\limits_{n\to\infty} na_n = 0$ when $(a_n)$ is monotone (careful: this is false when $(a_n)$ is not monotone). $\endgroup$ – kahen May 6 '13 at 21:03
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Hint

By index change prove: $$\sum_{k=1}^n x_k=\sum_{k=1}^n k(x_k-x_{k+1})+nx_{n+1}$$

then prove that the series $$\sum_{n=1}^\infty n(x_n-x_{n+1})$$ is convergent and deduce that the sequence $(nx_{n+1})$ is convergent to some $\lambda$.

Can $\lambda$ be different to zero? Why? and finaly how to prove that the sequence $(nx_n)$ converges also to zero?

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First, we have that $x_n \geq 0$ for all $n$. Now, consider a "Cauchy slice" with $n$ terms: $$ C_n\stackrel{\rm{}def}{=}\sum_{k=n+1}^{2n} x_k $$ What can you say about it? And by what can you lower bound it?

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