Telescoping sums to introduce integral and derivative relationship

Question

In order to introduce the fundamental theorem of Calculus to a 18 year-old class I was thinking about starting with simple telescoping sums (finite sums). It would be important to have examples that are really basic and don't need much algebra and notation to be developed. The most simple example I can think of is the Mengoli series $\sum \frac{1}{n(n+1)}$ (in some finite version) but maybe something more basic could be found that doesn't involve decomposition of algebraic fractions. Any idea?

Edit:

Connections between telescoping sums and FTC:

1. The telescoping sum formula is a discrete equivalent of the FTC where integral are replaced by sums and derivative by increments $a_{n+1}-a_n$: the formula $\sum_{k=0}^N(a_{n+1}-a_{n})=a_N-a_0$ has the same structure of $\int_a^bf'(t)dt=f(b)-f(a)$

2. Telescoping sums can be also used to prove the FTC by evaluating the integral $\int_a^bf'(t)dt$ with a Riemann sum and replacing $$ \sum_k f'(t_k) \Delta t \approx \sum_k [f(t_{k+1})-f(t_k)] =f(b) - f(a) $$

Answer 1

$\sum_{k=1}^N 2^{-k}=1-2^{-N}$

Answer 2

Exponentials seem the simplest way to go, let me start from Jessica B's answer and make it clearly about a telescoping sum (which it is as soon as one recall how to compute the sum of a geometric series!).

Let $a_n=\frac{1}{2}3^n$, for $n\in \mathbb{N}$. The discrete derivative of this sequence is given by $a'_n=a_{n+1}-a_n=3a_n-a_n= 3^n$. Now, the discrete integral of this new sequence is classically $$\sum_{n=0}^{N-1} a'_n = \sum_{n=0}^{N-1} 3^n = \frac{3^N-1}{3-1} = \frac{3^N-1}{2}$$ but this can be retrieved using the discrete analogue of FTC, aka telescoping sums: $$\sum_{n=0}^{N-1} a'_n= \sum_{n=0}^{N-1} \big(a_{n+1}-a_n \big)= a_N-a_0 = \frac{1}{2}3^N-\frac{1}{2}$$

The reason to use $3$ rather than $2$ as base is that with $a_n=2^n$, you get $a'_n=2^n$ too and this is confusing. This is also the occasion to explain in a different way the denominator in the sum of geometric series, and to recall that exponentials of base different from $e$ have a constant dropping of at differentiation. In some sense, $2$ is the natural discrete base while $e$ is the natural continuous base.

Answer 3

This is not more basic than your example, but you can generalize it by replacing $n+1$ by $n+k$. Using $$ \frac{1}{n (n+k)} = \frac{1}{k} \left( \frac{1}{n} - \frac{1}{n+k} \right) $$ one can derive $$ \sum_{n=1}^\infty \frac{1}{n (n+k)} = \frac{1}{k} \sum_{n=1}^k \frac{1}{n} = \frac{1}{k} H_k $$ which is certainly not obvious (to me).

For $k=1$, the infinite sum is $H_1/1=1$.

For $k=2$: $$ \frac{1}{2} \left[ \left( \frac{1}{1}-\frac{1}{3} \right) + \left( \frac{1}{2}-\frac{1}{4} \right) + \left( \frac{1}{3}-\frac{1}{5} \right) + \left( \frac{1}{4}-\frac{1}{6} \right) + \cdots \right] = \frac{1}{2} \left( 1 + \frac{1}{2} \right) $$ which is $H_2/2=(\frac{3}{2})/2=\frac{3}{4}$. In other words, everything cancels in the infinite sum except $\sum_1^k \frac{1}{n}$.