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

Question

I oversaw a high-school mathematics test the other day, and one of the problems was the following

Show, using induction or other means, that $$\sum_{i = 1}^n\frac1{i(i+1)} = 1-\frac1{n+1}$$

The induction proof is very standard, where the induction step relies on the fact that $\frac{1}{n+1} + \frac{1}{n(n+1)} = \frac{1}{n}$, and I'm sure it's been answered on this site before. However, I got intrigued by the "or other means" part of the question.

I don't know whether the teacher who wrote the test even considered any alternative solutions (he may just have written it so that if anyone has a crazy idea that works out, then they should get a full score for it), but I tried to find one. For instance, we may do the following telescope-ish argument: $$ \frac{1}{1(1+1)} + \frac{1}{2(2+1)} + \cdots + \frac{1}{(n-1)n} + \frac{1}{n(n+1)} + \frac{1}{n+1}\\ = \frac{1}{1(1+1)} + \frac{1}{2(2+1)} + \cdots + \frac{1}{(n-1)n} + \frac1n\\ \vdots\\ = \frac{1}{1(1+1)} + \frac12\\ = 1 $$ However, I feel that this is just an induction proof in disguise (hidden in the vertical dots). If one uses the mechanics of the induction proof to check whether the formula is true for a specific $n$, then one certainly does the exact same calculations as I have done here.

Is there a proof of this fact that clearly does not use induction (or at least hides it better)? The more elementary the better, and the ultimate goal would be to do it within the syllabus of the students taking the test (or at least not far from it). For reference, they should be familiar with the summation formula of arithmetic and geometric series and their derivations (so techniques resembling those would be well within specifications).

If there is a solution using calculus, then the students should be able to integrate elementary trigonometric functions, as well as exponential functions, logarithms and rational functions. They are familiar with integration by parts, substitution and partial fractions.

I welcome more advanced solutions as well, of course.

Answer 1

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.

Answer 2

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 $[1,n+1].$

Answer 3

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 $$ \int_0^\infty ds \left[e^{-s}-e^{-(n+1) s}\right]=\frac{n}{n+1}\ . $$

Answer 4

You can use finite difference method to solve this (although, it truly shines with more complex sums, and is an overkill here).

Define $(\Delta f)(x)$ as $f(x+1)-f(x)$ It's easy to see that operator $\Delta$ is linear. It's supposed to be similar to differential operator, and recall that $D(x^k) = kx^{k-1}$, here we have a similar fact, but not for standard power, but for falling factorial instead.

Define falling factorial of x to be $$(x)_n =\begin{cases}x(x-1)(x-2)\ldots(x-n+1)& n>0\\ 1 & n = 0\\ \frac{1}{(x+1)(x+2)\ldots(x+n)}& n < 0 \end{cases}$$

This operation is defined in such a way, to satisfy for all $n,m\in\mathbb{Z}$ the property $(x)_{m+n}=(x)_m(x-m)_n$
Moreover $\Delta((x)_n) = (x+1)_n - (x)_n = (x+1)x(x-1)\ldots(x-n+2) - x(x-1)\ldots(x-n+1) = x(x-1)\ldots(x-n+2)(x+1 - (x-n+1)) = n(x)_{n-1}$
I am only proving this property for positive $n$, but it holds in general.

Now, let's move one to sums. Define the $\Sigma$ operator to be
$$\sum g(x)\delta x = f(x) + p(x) \iff g(x) = \Delta f(x)$$ Where $p$ denotes a function satisfying $p(x+1)=p(x)$ (such functions vanish under $\Delta$ operator).

And let $\sum_a^b g(x)\delta x = f(b) - f(a)$ It is clear, that this new kind of sum is directly connected to standard summation - indeed $\sum_a^bg(x) \delta x = \sum_{k=a}^{b-1}g(k)$, which is obvious enough given that $\sum_a^a g(x) \delta x = f(a) - f(a) = 0$ and
$\sum_a^{b+1}g(x)\delta x - \sum_a^b g(x) \delta x = (f(b+1)-f(a)) - (f(b)-f(a)) = f(b+1) - f(b) = g(b)$

So, in our new notation, the sum in question can be rewritten as $$\sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{0}^{n} (x)_{-2}\delta x$$

Just like in calculus, remembering our remark about $\Delta$ of falling factorial we arrive at equation $$\sum (x)_n \delta x = \frac{(x)_{n+1}}{n+1} + C \hspace{10pt} \text{for } n \not= -1$$

So we have $\sum (x)_{-2}\delta x = -(x)_{-1} + C$ Plugging that into our formula we arrive at
$$\sum_{0}^{n}(x)_{-2}\delta x = -(n)_{-1} - (-(0)_{-1}) = 1- \frac{1}{n+1}$$