# Proof verification of if $\{x_n\}$ is monotone then $\{y_n\} = {1\over x_1 + x_2 + \dots + x_n}$ is monotone

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.

• Yes, it's wrong. For $x_n>0$ it's obvious. – Michael Rozenberg Nov 8 '18 at 13:01
• Inequality (3) has nothing to do with the fact that $\{x_i\}$ is a monotone sequence. It holds for any set of positive numbers – Oldboy Nov 8 '18 at 13:01

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!

• 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 $x_n$ – roman Nov 8 '18 at 13:13
• Yes, that's right. Even tha partial sums are not monotone! – Robert Z Nov 8 '18 at 13:14
• The problem makes sense onlu for $x_i\gt0$. – Oldboy Nov 8 '18 at 13:15
• Yes, if the $x$s have all the same sign then the monotone property is preserved. – Robert Z Nov 8 '18 at 13:17
• @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. – roman Nov 8 '18 at 13:18

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.

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}