Arithmetic Progression with a Sum Property Question

Question

I had a question in a exam, and I had no idea how to do it.

The question is:

Let $(a_k)_{k\geq1}$ a sequence of not null real numbers such that:

$$\displaystyle \sum_{k=1}^{n-1}\frac{1}{a_k\cdot a_{k+1}}=\frac{n-1}{a_1\cdot a_n} \, \forall \,n\geq 2$$

Show that this sequence is an arithmetic progression.

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I know that to show that this is an arithmetic progression, I have to show that $a_n-a_{n-1}$ is constant. But I don't know how to do it with the sum hypothesis.

Appreciate your help.

Answer 1

For any integer $n\ge2,$ from $$\frac1{a_na_{n+1}}=\sum_{k=1}^n\frac1{a_ka_{k+1}}-\sum_{k=1}^{n-1}\frac1{a_ka_{k+1}}=\frac n{a_1a_{n+1}}-\frac{n-1}{a_1a_n}$$we deduce: $$a_{n+1}=\frac{na_n-a_1}{n-1}.$$

Therefore, if $a_n=a_1+(n-1)r$ for some $r,$ then $a_{n+1}=\frac{n(a_1+(n-1)r)-a_1}{n-1}=a_1+nr$ for the same $r.$

Hence by induction, $\forall n\in\Bbb N\quad a_n=a_1+(n-1)(a_2-a_1).$

Answer 2

Note that: $$\frac{1}{a_{k}} - \frac{1}{a_{k+1}} = \frac{a_{k+1}-a_{k}}{a_{k+1}a_{k}} = \frac{d}{a_{k+1}a_{k}} = d(\frac{1}{a_{k+1}a_{k}})$$ if $d = a_{k+1} - a_{k}$. If you substitute every instance of $\frac{1}{a_{k+1}a_{k}}$ with $(\frac{1}{d})\frac{1}{a_{k+1}a_{k}}$, you can create a telescoping series: $\frac{1}{d}(\frac{1}{a_{1}} - \frac{1}{a_{2}} + \frac{1}{a_{2}} - \frac{1}{a_{3}}...)$ = $\frac{1}{d}(\frac{1}{a_{1}} - \frac{1}{a_{n}})$ = $\frac{1}{d}(\frac{a_{n} - a_{1}}{a_{n}a_{1}})$ = $\frac{1}{d}(\frac{(n-1)d}{a_{n}a_{1}})$ = $\frac{n-1}{a_{1}a_{n}}$

Answer 3

Let $r = a_2 - a_1$. To prove that $(a_n)$ is arithmetic is equivalent to prove that: $$\forall n \geq 1, a_n = (n - 1) r + a_1$$ Let's proceed by strong induction.

1. It's clear, it's true for $n = 1$.
2. Assume it's true for $k \in \{1, \ldots, n - 1\}$ then : $$\begin{array}{lcl} \displaystyle \dfrac{n - 1}{a_1 a_n} & = & \displaystyle \sum_{k = 1}^{n - 1} \dfrac{1}{a_k a_{k + 1}} \\[3mm] & = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \sum_{k = 1}^{n - 2} \dfrac{1}{a_k a_{k + 1}} \\[3mm] & = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \sum_{k = 1}^{n - 2} \dfrac{a_{k + 1} - a_k}{a_k a_{k + 1}} \\[3mm] & = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \sum_{k = 1}^{n - 2} \left(\dfrac{1}{a_k} - \dfrac{1}{a_{k + 1}}\right) \\[3mm] & = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \left(\dfrac{1}{a_1} - \dfrac{1}{a_{n - 1}}\right) \\[3mm] & = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \dfrac{a_{n - 1} - a_1}{a_1 a_{n - 1}} \\[3mm] & = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{1}{r} \dfrac{(n - 2) r + a_1 - a_1}{a_1 a_{n - 1}} \\[3mm] & = & \displaystyle \dfrac{1}{a_{n - 1} a_n} + \dfrac{n - 2}{a_1 a_{n - 1}} \end{array}$$ We deduce that : $$\begin{array}{lcl} \dfrac{n - 2}{a_1 a_{n - 1}} & = & \dfrac{1}{a_n} \left(\dfrac{n - 1}{a_1} - \dfrac{1}{a_{n - 1}}\right) \\[3mm] & = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) a_{n - 1} - a_1}{a_1 a_{n - 1}}\right) \\[3mm] & = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) ((n - 2) r + a_1) - a_1}{a_1 a_{n - 1}}\right) \\[3mm] & = & \dfrac{1}{a_n} \left(\dfrac{(n - 1) (n - 2) r + (n - 2) a_1}{a_1 a_{n - 1}}\right) \\[3mm] & = & \dfrac{n - 2}{a_n} \left(\dfrac{(n - 1) r + a_1}{a_1 a_{n - 1}}\right) \\[3mm] \end{array}$$ then : $$\dfrac{(n - 1) r + a_1}{a_n} = 1$$ so $a_n = (n - 1) r + a_1$.

By induction : $$\forall n \geq 1,a_n = (n - 1) r + a_1$$ so $(a_n)$ is arithmetic.