# Request for assistance on finishing the proof that $\lim_{n\to\infty}n x_n = 0$ given some initial conditions.

Let $$\{x_n\}$$ denote a sequence such that: $$\exists N \in \Bbb N: \forall n > N \implies 0 < x_{n+1} < x_n$$ And the sequence $$\{y_n\}$$ is convergent, where $$\{y_n\}$$ is given by: $$y_n = \sum_{k=1}^n x_n$$ Prove that: $$\lim_{n\to\infty} n x_n =0$$

Below is what I've done so far.

Start with the first fact. We are given that a sequence $$x_n$$ is monotonically decreasing starting from some $$N$$. That by Weierstrass theorem means that the sequence is convergent to some $$x_0 > 0$$, namely: $$\exists x_0 \in\Bbb R_{>0}:\lim_{n\to\infty} x_n = x_0 \tag1$$

Also it is bounded: $$\exists m, M \in \Bbb R,\ \forall n \in \Bbb N:m \le x_n \le M \tag2$$

We are also given that the sum is convergent, hence: $$\exists L \in \Bbb R:\lim_{n\to\infty} y_n = \lim_{n\to\infty}\sum_{k=1}^n x_k = L \tag3$$

Combining both facts above we may fix any $$\epsilon > 0$$ and find $$N$$ such that: $$\forall \epsilon > 0\ \exists N \in \Bbb N: \forall m, n > N \implies \begin{cases} |x_n| < \epsilon \\ |y_n - y_m| < \epsilon \end{cases}$$

By convergence we also have that $$\{y_n\}$$ is a Cauchy sequence.

Since both sequences converge we as well know that for $$\{x_n\}$$: $$\lim_{n\to\infty}\sup x_n = \lim_{n\to\infty}\inf x_n = x_0\tag4$$ And for $$y_n$$: $$\lim_{n\to\infty}\sup y_n = \lim_{n\to\infty}\inf y_n = L \tag5$$

I've been playing around with those properties for a long time yet, but I couldn't find a way to combine them in order to show that: $$\lim_{n\to\infty}nx_n = 0$$

Could someone please assist me on that? I would prefer a hint rather than a complete proof. Thank you!

• A monotonically decreasing sequence doesn't imply it tends to $0$. However, you could use the fact that $y_n$ convereges and hence $x_n$ must converge to $0$. Unless I'm missing something about Weierstrass Theorem. Which one is it? About 8 on wikipedia disambiguation page – Anvit Dec 27 '18 at 13:06
• $n < 2^n$ so Cauchy Condensation might work – Anvit Dec 27 '18 at 13:09
• @Anvit By this theorem I mean that any monotonic sequence $\{x_n\}$ has a finite limit in case it's bounded. – roman Dec 27 '18 at 13:11
• No, that stament alone doesn't work, define $z_n = 1 + x_n$. $z_n$ also satifies the above properties but converges to 1 (if $x_n$ converegs to $0$) – Anvit Dec 27 '18 at 13:14
• @Anvit is right when he says that $0 \lt x_{n+1} \lt x_n$ doesn't imply that $\lim_{n \to \infty} x_n = 0$. $x_n = 1 + \frac{1}{n}$ is a counter exemple. On the other hand, the fact that $y_n$ is convergent is a sufficient condition for $x_n$ convergence to $0$ – F.Carette Dec 27 '18 at 13:15

Hint: $$\lim_{n\to\inf} (y_{2n}-y_n) = 0$$
Comment if you require more hints

Also, as I said, you can also use Cauchy Condensation test as an Alternative solution

• Thank you, I've added an answer to this question, hopefully correct. – roman Dec 27 '18 at 14:21

My try to use hint by @Anvit

Lemma: $$\exists L \in \Bbb R: \lim_{n\to\infty} y_n= \lim_{n\to\infty}\sum_{k=1}^n x_n = L \implies \lim_{n\to\infty} x_n = 0$$

Proof. Since $$\{y_n\}$$ converges it must satisfy Cauchy Criterion. Consider the following: $$\forall \epsilon > 0\ \exists N \in \Bbb N: \forall n, m > N \implies |y_n -y_m| < \epsilon$$

Take $$m=n-1$$, then we have that: $$|x_n| = |y_n - y_{n-1}| < \epsilon$$ Which would mean: $$\forall \epsilon > 0\ \exists N \in\Bbb N: \forall n > N \implies |x_n| < \epsilon \stackrel{\text{def}}{\iff} \lim_{n\to\infty} x_n = 0$$

So by Lemma we obtain that $$x_n$$ is convergent to $$0$$: $$\lim_{n\to\infty}x_n = 0$$ As desired $$\Box$$.

Consider $$\{y_n\}$$ and $$m = 2n > n$$, then by Cauchy Criterion: \begin{align} |y_n - y_m| &= |y_m - y_n| \\ &= |y_{2n} - y_n| \\ &= \left| \sum_{k=n+1}^{2n}x_k \right| \\ &= \sum_{k=n+1}^{2n}x_k < \epsilon \end{align}

We know that $$x_n > 0$$ and is monotonically decreasing towards $$0$$ starting from some $$N$$ and thus: $$\underbrace{x_{2n} + x_{2n} + \cdots + x_{2n}}_{n\ \text{times}} < x_{n+1} + x_{n+2} + \cdots + x_{2n-1} + x_{2n} = \sum_{k=n+1}^{2n}x_k < \epsilon \\ nx_{2n} < \sum_{k=n+1}^{2n}x_k < \epsilon \\ 2n\cdot x_{2n} < 2\sum_{k=n+1}^{2n}x_k < 2\epsilon$$

We know that: $$\lim_{n\to\infty} 2\sum_{k=n+1}^{2n}x_k = 0$$

Applying squeeze theorem to that: $$0 \le \lim_{n\to\infty} 2n\cdot x_{2n} \le \lim_{n\to\infty} 2\sum_{k=n+1}^{2n}x_k$$

Using the fact that $$x_n$$ is monotonically decreasing we may conclude: $$0 \le \lim_{n\to\infty} 2n\cdot x_{2n} \le 0 \iff \\ \lim_{n\to\infty} 2n\cdot x_{2n} = 0 \iff \\ \lim_{n\to\infty} n\cdot x_{n} = 0$$

• Little problem I see is that your last iff statement isn't correct. Consider a sequence whose even numbers tend to zero but odds dont – Anvit Jan 7 at 16:01
• @Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something? – roman Jan 7 at 16:09
• Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P ) – Anvit Jan 7 at 16:11