I have to find the formula for the nth partial sum of this series in order to determine if the series converges or diverges. Here's the equation:
$$\sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+1})$$
What I did so far:
I made it into a single fraction:
$$\sum_{n=1}^\infty(\frac{(n+1)-n}{n(n+1)})$$
Then I split it up into partial fractions:
$$\frac{A}{n}-\frac{B}{n+1}=\frac{(n+1)-n}{n(n+1)}$$
$$\frac{An+A-Bn}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}$$
$$\frac{n(A-B)+A}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}$$
and so I got the two equations:
$$n(A+B)=-n$$
$$A=n+1$$
Does this look right so far? I know I'd have to use the Elimination method in order to find the sum.
Best Answer
$S_1=\frac{1}{1}-\frac{1}{1+1}=\frac{1}{2}$
$S_2=\frac{1}{1}-\frac{1}{1+1}+\frac{1}{2}-\frac{1}{2+1}=\frac{1}{1}-\frac{1}{3}$
$S_3=\frac{1}{1}-\frac{1}{1+1}+\frac{1}{2}-\frac{1}{2+1}+\frac{1}{3}-\frac{1}{3+1}=\frac{1}{1}-\frac{1}{3+1}$
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$S_n=\frac{1}{1}-\frac{1}{n+1}$