Proving the Sum of Reciprocals Formula using Induction

[wolfmanzak]

Homework Statement

Show that the following formula holds for any natural number n.

\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+ \cdots + \frac{1}{n(n + 1)}= \frac{n}{n+1}

Homework Equations

The Attempt at a Solution

I'm trying to decide the best way to do this problem. I would think trying Induction would be the better idea but I just wanted to get some opinions. I'm going to try to work on it some and post my attempt soon. In the meantime, I'd appreciate just some help with where to start or a general procedure to follow. Any and all help/solutions are welcome. Thank you in advance.

 

[hgfalling]

Uh yeah, induction is the way to go. It's very straightforward.

 

[wolfmanzak]

I have tried adding (k+1), 2(k+1), (k+1)^{2}...all of these to both sides of the equation and still cannot make it work. What do I need to add to the equation so that I can prove P(k+1)? Would I add something like \frac{1}{(k+1)((k+1)+1)}? I'm running out of ideas. I really need an answer to this problem.

 

[Office_Shredder]

\frac{1}{3*4} = \frac{1}{3} - \frac{1}{4}

 

[wolfmanzak]

Not sure I follow you. Are you saying that \frac{1}{k*(k+1)} = \frac{1}{k} - \frac{1}{k+1}? How would that help? Sorry, I just don't follow you at the moment.

 

[hgfalling]

Would I add something like \frac{1}{(k+1)((k+1)+1)}?

Yes, this.

Every time n increments, you add one more term, right? So the sum for k+1 is just the sum for k plus what you just wrote above.

 

[wolfmanzak]

Thanks, I can't believe I didn't see that. I've got it figured out now. I appreciate all of the help.