# Telescoping sums to introduce integral and derivative relationship

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 $$\sum_k f'(t_k) \Delta t \approx \sum_k [f(t_{k+1})-f(t_k)] =f(b) - f(a)$$

• Can you outline the method? I'm interested in alternative methods for introducing the FTC. May 4, 2016 at 13:47
• I start showing an example of a sequence of numbers I want to sum, and do the sum, then I reveal to them that these numbers were obtained as increments $a_{n+1}-a_{n}$ from another sequence, and then it is easy to realize that we didn't need to compute all the sumation: instead we could just compute the difference $a_{\text{last}}-a_0$ to get the same result more quickly. May 4, 2016 at 14:08
• I don't see the connection between the Fundamental Theorem of Calculus and a telescoping series. Perhaps you could show how the two are connected. I warn you, if it's not immediately obvious to someone like me, or someone else in math, how will it be clear to someone learning the material? May 5, 2016 at 2:57
• I also fear that you are not really talking about the "Fundamental Theorem of Calculus" which states that for any integral $\int_a^b f(x)dx$, such that $f(x)$ is continuous, there exists a function $F(x)$ such that $F'(x) = f(x)$ and $\int_a^b f(x)dx = F(b) - F(a)$. I think what you are talking about, possibly (I don't fully understand), is showing that perhaps an antiderivative can be used to evaluate an integral--not that an antiderivative can be used to evaluate a definite integral. May 5, 2016 at 3:03
• There are several 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 $a_{n+1}-a_n$ ($\sum_{k=0}^N(a_{n+1}-a_{n})=a_N-a_0$ has the same structure of $\int_a^bf'(t)dt=f(b)-f(a)$) (2) Telescoping sums can be used to prove the FTC evaluating the integral by a Riemann sum and replacing $f'(t_k) \Delta t \approx f(t_{k+1})-f(t_k)$. May 5, 2016 at 14:29

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

• How do you see it as a telescoping sum? May 4, 2016 at 12:51
• @MarcoDisce I don't know what sort of answer you want. If you want the terms of the sum to cancel, then you have something of the form $f(n)-f(n-1)$. If the terms appear in the two parts, then it doesn't much matter what you use for $f$, within what the students are reasonably familiar with. If you don't want the two parts to be immediately visible, some sort of algebraic manipulation is going to be needed. May 4, 2016 at 20:30
• @JessicaB Same comment as below, I don't see the connection to the Fundamental Theorem of Calculus. May 5, 2016 at 3:08
• @Jared The question isn't about the FTC. May 5, 2016 at 7:00
• @JessicaB The OP is saying they are using telescoping series to introduce the FTC. And actually the OP gave a good comment of their own on how telescoping series are relevant to the FTC. May 6, 2016 at 0:06

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.

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 $\sum_1^k \frac{1}{n}$.

• I don't see the connection to the Fundamental Theorem of Calculus...can you elaborate? May 5, 2016 at 3:07
• @Jared: I await the OP's connection to the FTC. I was only responding to the request for examples of telescoping series. May 5, 2016 at 9:56