# Prove $x_n = \sum_{k=1}^n \frac{1}{(a+(k-1)\cdot d)\cdot(a+k\cdot d)}$ is a bounded sequence.

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.

For $$d=0$$ it's not bounded, of course.
For $$d\neq0$$ and $$a+kd\neq0$$ use $$\sum_{k=1}^n\frac{1}{(a+(k-1)d)(a+kd)}=\frac{1}{d}\sum_{k=1}^n\left(\frac{1}{a+(k-1)d}-\frac{1}{a+kd}\right)=\frac{1}{d}\left(\frac{1}{a}-\frac{1}{a+nd}\right)$$