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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!

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  • $\begingroup$ 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 $\endgroup$ – Anvit Dec 27 '18 at 13:06
  • $\begingroup$ $n < 2^n$ so Cauchy Condensation might work $\endgroup$ – Anvit Dec 27 '18 at 13:09
  • $\begingroup$ @Anvit By this theorem I mean that any monotonic sequence $\{x_n\}$ has a finite limit in case it's bounded. $\endgroup$ – roman Dec 27 '18 at 13:11
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    $\begingroup$ 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 $0$) $\endgroup$ – Anvit Dec 27 '18 at 13:14
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    $\begingroup$ @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 $0$ $\endgroup$ – F.Carette Dec 27 '18 at 13:15
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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

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  • $\begingroup$ Thank you, I've added an answer to this question, hopefully correct. $\endgroup$ – roman Dec 27 '18 at 14:21
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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: $$ 0 \le \lim_{n\to\infty} 2n\cdot x_{2n} \le 0 \iff \\ \lim_{n\to\infty} 2n\cdot x_{2n} = 0 \iff \\ \lim_{n\to\infty} n\cdot x_{n} = 0 $$

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  • $\begingroup$ 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 $\endgroup$ – Anvit Jan 7 at 16:01
  • $\begingroup$ @Anvit But then the sequence is not going to be monotone, is it? which violates the initial conditions, or am i missing something? $\endgroup$ – roman Jan 7 at 16:09
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    $\begingroup$ Yep, you're right. However you should mention this else it can lead to confusion. (And possible loss of marks :P ) $\endgroup$ – Anvit Jan 7 at 16:11

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