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Let $n\in \mathbb N$ and $\{x_n\}$ is a monotone sequence. Prove that: $$ \{y_n\} = {1\over x_1 + x_2 + \dots + x_n} $$ is also a monotone sequence.

Given $\{x_n\}$ is monotone then by definition: $$ \forall n\in\mathbb N:x_n \le x_{n+1} \tag1 $$ or: $$ \forall n\in\mathbb N:x_n \ge x_{n+1} \tag2 $$ Let's prove for $(1)$ first. Using definition of a monotone sequence: $$ x_1 \le x_2 \le x_3 \le \dots \le x_n $$ So from this: $$ x_1 + x_2 + x_3 + \dots + x_n \le x_1 + x_2 + x_3 + \dots + x_n + x_{n+1} \tag 3 $$ Thus taking the reciprocal of $(3)$ we get that:

$$ {1\over x_1 + x_2 + x_3 + \dots + x_n} \ge {1\over x_1 + x_2 + x_3 + \dots + x_n + x_{n+1}} $$ But LHS is $y_n$ and RHS of the inequality is $y_{n+1}$, therefore by definition of a monotone sequence we conclude that $y_n$ is also monotone.

Case $(2)$ is obtained similarly. What bugs me is that the following still holds (at least for $x_n \ge 0$):

$$ x_1 + x_2 + x_3 + \dots + x_n \le x_1 + x_2 + x_3 + \dots + x_n + x_{n+1} $$

And we get once again that:

$$ {1\over x_1 + x_2 + x_3 + \dots + x_n} \ge {1\over x_1 + x_2 + x_3 + \dots + x_n + x_{n+1}} $$

But how is this possible? From the above it looks like $y_n$ is always monotonically decreasing, which seems false. For example when $\sum_{k=1}^nx_k < 1$.

Where have i taken the wrong road?

Update

As shown in answers and comments monotonicity is only preserved assuming all $x_n$ have the same sign.

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    $\begingroup$ Yes, it's wrong. For $x_n>0$ it's obvious. $\endgroup$ – Michael Rozenberg Nov 8 '18 at 13:01
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    $\begingroup$ Inequality (3) has nothing to do with the fact that $\{x_i\}$ is a monotone sequence. It holds for any set of positive numbers $\endgroup$ – Oldboy Nov 8 '18 at 13:01
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The sequence $-3,-2,-1,0,1,2,4,5,\dots$ is monotone, the sequence of the partial sums $-3,-5,-6,-6,-5,-3,1,6,\dots$ is not monotone, and the sequence of the reciprocals of the partial sums is still not monotone.

For such sequences the monotone property is preserved if the $x$s have all the same sign!

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  • $\begingroup$ So $x_n$ is monotone does not imply $y_n$ is monotone, right? And hence such a proof may not be obtained, at least without additional assumptions on $x_n$ $\endgroup$ – roman Nov 8 '18 at 13:13
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    $\begingroup$ Yes, that's right. Even tha partial sums are not monotone! $\endgroup$ – Robert Z Nov 8 '18 at 13:14
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    $\begingroup$ The problem makes sense onlu for $x_i\gt0$. $\endgroup$ – Oldboy Nov 8 '18 at 13:15
  • $\begingroup$ Yes, if the $x$s have all the same sign then the monotone property is preserved. $\endgroup$ – Robert Z Nov 8 '18 at 13:17
  • $\begingroup$ @Oldboy well I agree with you, though the author didn't put in any constraints for $x_n$. I'm going to stick to $x_n > 0$. Otherwise it makes little sense. $\endgroup$ – roman Nov 8 '18 at 13:18
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You mistakenly assume (in case (1) and in case (2) alike) that $x\le y$ implies $\frac1x\ge\frac1y$, whereas this is the case only if $xy>0$. Consider the sequence $x_1=-6$, $x_2=2$, $x_3=3$, $x_4=4$, and then whatever.

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For $x_i>0$, it's trivial:

$$y_n = {1\over x_1 + x_2 + \dots + x_n}$$

$$y_{n+1} = {1\over x_1 + x_2 + \dots + x_n + x_{n+1}}$$

$$y_{n+1} -y_n= {1\over x_1 + x_2 + \dots + x_n + x_{n+1}}-{1\over x_1 + x_2 + \dots + x_n}$$

$$y_{n+1} -y_n= {-x_{n+1} \over (x_1 + x_2 + \dots + x_n + x_{n+1})(x_1 + x_2 + \dots + x_n)}<0$$

$$y_{n+1} <y_n$$

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