Hey,
HallsofIvy said:
<br />
\sum_{i=1}^{100} \frac{1}{i(i+1}= \frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100} = (\frac{1}{1}- \frac{1}{2})+ (\frac{1}{2}-\frac{1}{3})+ (\frac{1}{3}-\frac{1}{4})+ \cdot\cdot\cdot+ (\frac{1}{99}- \frac{1}{100})<br />
The last fraction in each pair cancels the first fraction in the next pair (except of course in the last pair). Every fraction except the first and last cancel so the sum is
\frac{1}{1}- \frac{1}{100}= \frac{100-1}{100}= \frac{99}{100}
More generally,
\sum_{i= 1}^n \frac{1}{i(i+1)}= \frac{1}{1}- \frac{1}{n+1}= \frac{n+1-1}{n+1}= \frac{n}{n+1}
Thanks HallsofIvy for showing how it could have been proved by showing it was a telescoping sum.
However, how would you know that the individual sums can be rewritten as the sum or difference of two fractions. In other words, how did you figure out the following,
<br />
\sum_{i=1}^{100} \frac{1}{i(i+1}= \frac{1}{1\cdot2}} + {\frac{1}{2\cdot3}} + {\frac{1}{3\cdot4}} + . . . + {\frac{1}{99\cdot100} = \left(\frac{1}{1}-\frac{1}{2}\right) + \left(\frac{1}{2}-\frac{1}{3}\right)+ \left(\frac{1}{3}-\frac{1}{4}\right) + \cdot\cdot\cdot + \left(\frac{1}{99}- \frac{1}{100}\right)<br />
That is, how did you figure out that,
<br />
{{\sum_{i = 1}^{n}}{\frac{1}{i(i+1)}}} = {\left({{\frac{A_{1}}{1}}-{\frac{B_{1}}{2}}}\right)} + {\left({{\frac{A_{2}}{2}}-{\frac{B_{2}}{3}}}\right)} +. . . + {\left({{\frac{A_{n}}{n}}-{\frac{B_{n}}{n+1}}}\right)}<br />
Additionally, is finding the above the same as using the technique of partial fractions?
Also, what is the general way of finding A and C for the following (where: B, D, E, and{\textcolor[rgb]{1.00,1.00,1.00}{.}}F; are all given),
<br />
{\frac{E}{F}} = {{{\frac{A}{B}}\pm{\frac{C}{D}}}}<br />
Where it can be shown that,
<br />
{E} = {AD \pm BC}<br />
<br />
{F} = {BD}<br />
Which is the same as,
<br />
{\frac{E}{F}} = {\frac{AD \pm BC}{BD}}<br />
Thanks,
-PFStudent