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Let $n \in \mathbb N$ and: $$ x_n = \sum_{k=1}^n \frac{1}{(a+(k-1)\cdot d)\cdot(a+k\cdot d)} $$ Prove $\{x_n\}$ is a bounded sequence.

I'm having hard time finishing the proof. Below is what i've done so far.

I've started with inspecting $x_1, x_2, x_3 \dots$:

$$ \begin{align} x_1 &= \frac{1}{a\cdot(a+d)} \\ x_2 &= x_1 + \frac{1}{(a+d)\cdot(a+2d)} = \frac{2}{a\cdot(a+2d)} \\ x_3 &= x_2 + \frac{1}{a(a+2d)\cdot(a+3d)} = \frac{3}{a\cdot(a+3d)} \\ &\dots \\ x_n &= \frac{n}{a\cdot(a+n\cdot d)} \end{align} $$

I've shown by induction that this holds for $x_{n+1}$.

Please note that no information about $a, d$ is given in the problem statement. So for simplicity I'm going to assume they are just positive integers.

So now:

$$ \begin{cases} x_n = \frac{n}{a\cdot(a+n\cdot d)} \\ a, d \in \mathbb N \end{cases} $$

Clearly from this point I could use $\lim_{n \to \infty} x_n$ but I'm not allowed to use limits. Each term of $x_n$ is decreasing (at least with assumptions for $a, d$) therefore:

$$ \frac{1}{a\cdot d + a^2} \le x_n < {1 \over a\cdot d} $$

Hence the sequence is bounded. What i'm struggling with is how to show $\{x_n\}$ is bounded without using a limit.

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For $d=0$ it's not bounded, of course.

For $d\neq0$ and $a+kd\neq0$ use $$\sum_{k=1}^n\frac{1}{(a+(k-1)d)(a+kd)}=\frac{1}{d}\sum_{k=1}^n\left(\frac{1}{a+(k-1)d}-\frac{1}{a+kd}\right)=\frac{1}{d}\left(\frac{1}{a}-\frac{1}{a+nd}\right)$$

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  • $\begingroup$ Thats so obvious now, i should have split the sum input partial fractions right away. Thank you! $\endgroup$ – roman Oct 22 '18 at 17:50
  • $\begingroup$ You are welcome! $\endgroup$ – Michael Rozenberg Oct 22 '18 at 17:55

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