# Help in understanding this...

On the wiki page for telescoping series, this example is given....

Can someone help me understand the second step, where the fraction is reduced to all 1s in the numerator? I've researched the internet to understand how this was so simply done. How does (4-1)/(1*4) reduce so nicely? (I know it does, obviously), but there is a template here where the numerators go to 1 while the denominators go individually to, say, 1 and 4 (in the first fraction).

Is there a reason 1/3 was pulled out of the series? How was that number derived, other than trial and error.

It a general rule that (x-y)/(yx) can be reduced to 1/y minus 1/x. Is creating a situation where you have (x-y)/(yx) the basis for finding the 1/3 that is outside the parentheses and for how you go about solving this type of problem more generally?

6 months, 1 week ago

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Check out Partial Fractions to learn how to decompose such fractions.

Staff - 6 months, 1 week ago

This follows by simply observing that

$\frac{4-1}{1\cdot4}=\frac{4}{1\cdot4}-\frac{1}{1\cdot4}=\frac{1}{1}-\frac{1}{4}$

The reason for taking out $\displaystyle \frac{1}{3}$ is for the numerators to get cancelled out...….here's an example...…

$\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+...$

$=\frac{1}{6}\left(\frac{6}{1\cdot7}+\frac{6}{7\cdot13}+\frac{6}{13\cdot19}+...\right)$

$=\frac{1}{6}\left(\frac{7-1}{1\cdot7}+\frac{13-7}{7\cdot13}+\frac{19-13}{13\cdot19}+...\right)$

$=\frac{1}{6}\left(\frac{7}{1\cdot7}-\frac{1}{1\cdot7}+\frac{13}{7\cdot13}-\frac{7}{7\cdot13}+\frac{19}{13\cdot19}-\frac{13}{13\cdot19}+...\right)$

$=\frac{1}{6}\left(\frac{1}{1}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{19}+...\right)$

$=\frac{1}{6}$

- 6 months, 1 week ago