Proof by induction: summation inductive step

Question

Disclaimer: This question is just a practice question and is not for marks.

I am trying to prove the following statement (I'm skipping right to the inductive step here since the base case is trivial):

$\forall n \in \mathbb{N}, \hspace{3pt} $$\sum_{i=1}^{n} \frac{1}{i(i+1)} = \frac{n}{n+1}$

Inductive Step: Let P(k) be $\forall k \in \mathbb{N}, \hspace{3pt} $$\sum_{i=1}^{k} \frac{1}{i(i+1)} = \frac{k}{k+1}$. We will prove $\forall k \in \mathbb{N}, \hspace{3pt} $$\sum_{i=1}^{k+1} \frac{1}{i(i+1)} = \frac{k+1}{k+2}$ is true.

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

I'm not sure whether the above step is correct, or how to progress with this proof. Also, is there a cookie cutter proof structure for this type of question? Thanks in advance for the assistance.

The part with $\frac{1}{k(k+1)}$ is wrong. When you replaced the $i$ in $\frac{1}{i(i+1)}$ with $k+1$, that's not quite what you should get. — balddraz, Apr 07 '19 at 02:49
Should it be $(k +1)$ instead? Edit: Oh, now I see, my substitution was wrong. — Shrek.Spongebob, Apr 07 '19 at 02:52
Yes, you have to replace the $i$ with $k + 1$ there. What do you get then? — balddraz, Apr 07 '19 at 02:55

score 1 · Accepted Answer · answered Apr 07 '19 at 02:58

1

Note that $$\sum_{i=1}^{k+1} \frac{1}{i(i+1)} = \sum_{i=1}^{k} \frac{1}{i(i+1)} + \frac{1}{(k+1)(k+2)}\\=\frac {k}{k+1}+\frac{1}{(k+1)(k+2)}.$$

As an alternative, observe that $$\sum_{i=1}^{k} \frac{1}{i(i+1)}=\sum_{i=1}^{k}\frac1i-\frac1 {i+1} $$.

answered Apr 07 '19 at 02:58

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Proof by induction: summation inductive step

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