proof by induction - algebra mistake?

Question

I have been working on this proof for a few hours and I can not make it work out.

$$\sum_{i=1}^{n}\frac{1}{i(i+1)}=1-\frac{1}{(n+1)}$$

i need to get to $1-\frac{1}{k+2}$

I get as far as $$1-\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}$$ then I have tried $1-\frac{(k+2)+1}{(k+1)(k+2)}$ which got me exactly nowhere.

score 1 · Answer 1 · answered Mar 05 '14 at 06:22

To finish your method:

We have $1 - \frac{1}{k+1} + \frac{1}{(k+1)(k+2)}$. You made a slight algebraic mistake -- you didn't distribute the negative when you combined fractions. If you did, you would have gotten $1 - \frac{(k+2) - 1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}$, thus completing the proof.

score 0 · Accepted Answer · answered Mar 05 '14 at 06:11

0

Hint: Write $\frac 1{i(i+1)}$ as $\frac 1i - \frac 1{i+1}$. Then you have a telescoping sum.

answered Mar 05 '14 at 06:11

ljfa

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omg... what a waste of so many hours lol.. that made it soo easy – Mikky Davey Mar 05 '14 at 06:18
It is easier that way than by induction :) – ljfa Mar 05 '14 at 06:20

score 0 · Answer 3 · answered Mar 05 '14 at 06:15

Claim $\sum_{n=1}^N \frac{1}{N(N+1)}=1-\frac{1}{N+1}$

If $N=0$, so $\sum_{n=1}^0 1/n-1/(n+1)=0$ and $1-1/(N+1)=0$. Suppose we have proven the assertion for $N\ge0$. Then

\begin{align}\sum_{n=1}^{N+1} 1/n-1/(n+1)=1/(N+1)-1/(N+2)+\sum_{n=1}^N 1/n-1/(n+1)\\ =1/(N+1)-1/(N+2)+1-1/(N+1)\\ =1-1/(N+2)\end{align}

proof by induction - algebra mistake?

3 Answers3