# Prove that $u_n$ is arithmetic sequence if $\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+\cdots+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$

Let $(u_n)$ be a sequence $u_i\neq0$ and $$\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+\cdots+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$$ for all $n\geq3$ Prove that the sequence $(u_n)$ is arithmetic sequence

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If this is a homework, then please add tag (homework). Also, it is quite recommended that you include, in the question, a log of your attempts, and where do you get stuck. –  user2468 Apr 3 '12 at 23:21
I would start by making a spreadsheet of 10 to 20 terms, try some $u_1$'s and $u_2$'s and see if you get a hint from that. –  Ross Millikan Apr 3 '12 at 23:30

## 3 Answers

There are cleverer ways of doing it, but here’s how you might approach it quasi-experimentally and arrive at a workable idea.

Start by looking at $n=3$: you have $$\frac1{u_1u_2}+\frac1{u_2u_3}=\frac2{u_1u_3}\;,$$ so $$\frac{u_1+u_3}{u_1u_2u_3}=\frac2{u_1u_3}\;,\tag{1}$$ and therefore $u_1+u_3=2u_2$. This says that $u_2$ is the mean of $u_1$ and $u_3$, so it’s midway between $u_1$ and $u_3$, and the first three terms therefore do indeed form an arithmetic progression. Let $d=u_2-u_1$.

What happens when $n=4$? You have $$\frac1{u_1u_2}+\frac1{u_2u_3}+\frac1{u_3u_4}=\frac3{u_1u_4}\;,$$ so $$\frac{u_3u_4+u_1u_4+u_1u_2}{u_1u_2u_3u_4}=\frac{3u_2u_3}{u_1u_2u_3u_4}\;,\tag{2}$$ and $$u_1u_2+u_3u_4+u_1u_4=3u_2u_3\;.\tag{3}$$ This implies that \begin{align*} u_4&=\frac{3u_2u_3-u_1u_2}{u_1+u_3}\tag{4}\\ &=\frac{u_2(3u_3-u_1)}{u_1+u_3}\\ &=\frac{(u_1+d)(2u_1+6d)}{2u_1+2d}\\\\ &=u_1+3d\;, \end{align*}

exactly what’s needed to keep it in arithmetic progression with the first three terms.

One more: $n=5$. The same kind of calculation leads to the equation $$u_5=\frac{4u_2u_3u_4-u_1u_2u_3}{u_1u_2+u_3u_4+u_1u_4}\;.\tag{5}$$ We already know from $(2)$ that the denominator of $(4)$ is $3u_2u_3$, so $$u_5=\frac{u_2u_3(4u_4-u_1)}{3u_2u_3}=\frac{3u_1+12d}3=u_1+4d\;,$$ as desired.

Now observe that the denominator in $(4)$ is the numerator in $(1)$, and the denominator in $(5)$ is the numerator in $(2)$. This should give you enough clues to pursue a proof by induction on $n$.

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We have

$$\rm s_n:=\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}.$$

Thus,

$$\rm s_{n+1}-s_n=\frac{1}{u_n u_{n+1}}=\frac{1}{u_1}\left(\frac{n}{u_{n+1}}-\frac{n-1}{u_n}\right).$$

Multiply both sides by $\rm u_1u_nu_{n+1}$, negate, add $\rm u_{n+1}$ to both sides,

$$\rm u_{n+1}-u_1=n(u_{n+1}-u_n).$$

Suppose, as our induction hypothesis, that $\rm u_k=a_1+d(k-1)$ for $\rm k=1,2,\cdots,n$ and some $\rm d$. Then the equation above is linear in $\rm u_{n+1}$, thus has a unique solution, and plugging in $\rm u_{n+1}=a_1+nd$ clearly works so it must be precisely that solution. The base case $\rm a_1=a_1$ is clear and examining the next case $\rm n=2$ tells us $\rm d=a_2-a_1$.

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What does "negate" mean? –  Pedro Tamaroff Apr 4 '12 at 0:56
@Peter Multiply by negative one. –  anon Apr 4 '12 at 0:57
Oh. It reminds me of old math papers. Do you think you could help me polish my argument? –  Pedro Tamaroff Apr 4 '12 at 1:01
Sure, I'll take a look at it. –  anon Apr 4 '12 at 1:04

You need two key features of an arithmetic progression:

$$a_n = a_1 +d(n-1)$$

and

$$a_n-a_{n-1}=d$$ (which is a consequence of the previous one).

Thus

$$\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n-1}u_n}=\frac{n-1}{u_1u_n}$$

$$d\left(\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1}{u_3u_4}+....+\frac{1}{u_{n- 1}u_n}\right)=\frac{d(n-1)}{u_1u_n}$$

Now sum $\dfrac{u_1}{u_1 u_n}$ to get

$$\frac{u_1}{u_1 u_n} + d\left(\frac{1}{u_1u_2}+\frac{1}{u_2u_3}+\frac{1} {u_3u_4}+....+\frac{1}{u_{n-1}u_n}\right)=\frac{u_1+d(n-1)}{u_1u_n}$$

$$\frac{1}{u_n} + \frac{d}{u_1u_2}+\frac{d}{u_2u_3}+\frac{d}{u_3u_4}+....+\frac{d}{u_{n-1}u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$

Now replace $d$ by the differences, conveniently:

$$\frac{1}{u_n} + \frac{u_2-u_1}{u_1u_2}+\frac{u_3-u_2}{u_2u_3}+\frac{u_4-u_3}{u_3u_4}+....+\frac{u_n-u_{n-1}}{u_{n-1}u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$

$$\frac{1}{u_n} + \frac{1}{u_1}-\frac{1}{u_2}+\frac{1}{u_2}-\frac{1}{u_3}+-....+\frac{1}{u_{n-1}}-\frac{1}{u_n}=\frac{u_1+d(n-1)}{u_1u_n}$$

This telescopes, giving

$$\frac{1}{u_1}=\frac{u_1+d(n-1)}{u_1u_n}$$

or

$$u_n=u_1+d(n-1)$$

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This can be turned into a legitimate argument, but it’s not one as it stands. It needs to be recast as an argument by induction on $n$, and the step replacing $u_1+d(n-1)$ by $u_n$ is invalid: it assumes the desired conclusion. What you should have, after telescoping, is $\frac1{u_1}=\frac{u_1+d(n-1)}{u_1u_n}$, from which the desired $u_n=u_1+d(n-1)$ follows immediately. –  Brian M. Scott Apr 4 '12 at 0:24
@Brian You're right. I was considering the ($\Leftarrow$) part of the argument (working it backwards). That's why I multiplied by the supposed $d$, and assumed $u_n-u_{n-1}=d$. –  Pedro Tamaroff Apr 4 '12 at 0:47
This is indeed the converse of the desired conclusion, but it can be made to work as a valid argument using a uniqueness lemma. You've shown the arithmetic sequence beginning with $u_1$, $u_2$ will satisfy the given recursion, and any sequence satisfying the given recursion is determined uniquely by its first two terms (see how?). Thus, the given sequence and the arithmetic sequence beginning with $u_1$ and $u_2$ must be one and the same. –  anon Apr 4 '12 at 1:11
@anon I see! Thanks. –  Pedro Tamaroff Apr 4 '12 at 1:12