Telescoping series seems not telescoping

Question

In the following example of the book Partial Differential Equation, An Introduction 2nd edition from Strauss, on page 127, they assert the following:

Let $f_n(x) = (1-x)x^{n-1}$ on the interval $ 0 < x < 1$. Then the series is telescoping. The partial sums are \begin{equation} \sum_{i = 1}^N f_n(x) = 1 - x^N \end{equation} Why does this series telescope? Computing partial sums does not yield cancellations.

Answer 1

$\require{cancel}$Note that $f_n(x)=x^{n-1}-x^n$ and that therefore$$\sum_{i=1}^Nf_n(x)=1-\cancel x+\cancel x-\cancel{x^2}+\cancel{x^2}-\cdots-\cancel{x^{N-1}}+\cancel{x^{N-1}}-x^N.$$So, yes, it is telescoping.

Answer 2

Yes, this is a telescopic sum! Note that $$\sum_{i = 1}^N f_n(x)=\sum_{i = 1}^N (x^{n-1}-x^n)=(1-x)+(x-x^2)+\dots+(x^{N-1}-x^N)=1-x^N.$$

Answer 3

Strauss clearly writes in the book (page 126) you mentioned:

Let $f_n(x)=(1-x)x^{n-1}$ on the interval $0<x<1$. Then the series is "telescoping." The partial sums are $$ \sum_1^Nf_n(x)=\sum_1^N(x^{n-1}-x^n)=1-x^N\to 1\quad \textrm{as }N\to\infty. $$

Here is the original excerpt from the book.