Showing $\sum_{i = 1}^n\frac1{i(i+1)} = 1-\frac1{n+1}$ without induction?

You can use a partial fractions decomposition, which assumes that we can rewrite $$ \frac{1}{i(i+1)} = \frac{A}{i} + \frac{B}{i+1}, \qquad 1 \leq i \leq n $$ for some $A, B$. Clearing denominators, the above equation is $1 = A(i+1) + B i$ for $1 \leq i \leq n$ which has solution $A=1$, $B=-1$. Hence \begin{align*} \sum_{i=1}^n \frac{1}{i(i+1)} &= \sum_{i=1}^n \frac{1}{i} - \frac{1}{i+1} \\ &= \left(1 \color{red}{- \frac{1}{2}} \right) + \left(\color{red}{\frac{1}{2}} \color{orange}{- \frac{1}{3}} \right) + \left(\color{orange}{\frac{1}{3}} \color{green}{- \frac{1}{4}} \right) + \cdots + \left(\color{blue}{\frac{1}{n-1}} \color{purple}{- \frac{1}{n}} \right) + \left( \color{purple}{\frac{1}{n}} - \frac{1}{n+1} \right) \\ &= 1 - \frac{1}{n+1}. \end{align*}

All adjacent (colored) summands cancel each other except for $1$ and $-\frac{1}{n+1}$, so we're left with $1 - \frac{1}{n+1}$. There's nothing inductive about it. It's just a convenient finite sum.

For a proof of intermediate level, note that for each $k=1,2....,n$ we have $$\int_k^{k+1} \frac{1}{t^2}\ dt=\frac{1}{k(k+1)}.$$ Adding these integrals for $k$ from $1$ to $n$ then gives $$\int_1^{n+1}\frac{1}{t^2}\ dt=1-\frac{1}{n+1}.$$

In this approach the "telescoping" may be said to hide in the additivity of the integral over the disjoint subintervals of $[1,n+1].$

You can write $$ \sum_{i = 1}^n\frac1{i(i+1)} = \sum_{i = 1}^n\left[\frac{1}{i}-\frac{1}{i+1}\right]\ . $$ Then use the identity $$ \frac{1}{x}=\int_0^\infty ds e^{-s x}\qquad x>0\ , $$ to write $$ \sum_{i = 1}^n\left[\frac{1}{i}-\frac{1}{i+1}\right]=\int_0^\infty ds\sum_{i=1}^n \left[e^{-s i}-e^{-s (i+1)}\right]\ , $$ and then use geometric sums $$ \int_0^\infty ds \left[e^{-s}-e^{-(n+1) s}\right]=\frac{n}{n+1}\ . $$