# How to find formula for nth partial sum of Telescopic Series

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.

• what is the partial sum for $n$ up to $2?$ What about up to $3?$ – Will Jagy Oct 10 '18 at 0:06
• Please don't make it into one fraction. Just add it as two fractions. Then it's pretty! – Don Thousand Oct 10 '18 at 0:07
• it's called "telescopic" for a reason. I have a feeling that a lot of fractions will cancel each other out and you'll only have the first and the last one left – Vasya Oct 10 '18 at 0:12
• Take a step back from the board: obviously $A=B=1$, that's where you started from. – Arnaud Mortier Oct 10 '18 at 0:17

$$S_1=\frac{1}{1}-\frac{1}{1+1}=\frac{1}{2}$$

$$S_2=\frac{1}{1}-\frac{1}{1+1}+\frac{1}{2}-\frac{1}{2+1}=\frac{1}{1}-\frac{1}{3}$$

$$S_3=\frac{1}{1}-\frac{1}{1+1}+\frac{1}{2}-\frac{1}{2+1}+\frac{1}{3}-\frac{1}{3+1}=\frac{1}{1}-\frac{1}{3+1}$$

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$$S_n=\frac{1}{1}-\frac{1}{n+1}$$