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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.

Maurice
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3 Answers3

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$\require{cancel}$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.

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

Robert Z
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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.

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  • I did not find the identity $\sum f_n(x) = \sum_1^N (x^{n-1} - x^n)$ – Maurice Jun 27 '19 at 13:57
  • @Rice4000 Which edition are you reading? You should probably find the newest edition. –  Jun 27 '19 at 13:57
  • @Rice4000 Compare your text with http://jingweizhu.weebly.com/uploads/1/3/5/4/13548262/partial-differential-equations.pdf – Robert Z Jun 27 '19 at 14:01
  • @Rice4000: I'm surprised you do not find that identity in Strauss's book since you have mentioned "edition 2" in your post. –  Jun 27 '19 at 14:05