Mathematical Induction - An example

Hi there!

Just working out one example but I'm having problems ...

Given:

{S}_n = \frac{n}{n+1}

Step 1:

{n} = {k}

{S}_k = \frac{1}{k(k+1)} = \frac{k}{k+1}

Step 2:

{n} = {k+1}

{S}_{k+1} = \frac{k+1}{k+2}

So far, so good.

What I do not understand is the next step:

Step 3:

{S}_{k+1} = {S}_{k} + \frac{1}{(k+1)(k+2)}

Where do I get that (

\frac{1}{(k+1)(k+2)}

)

Thanks in advance!

Reply 1

generalebriety

Can you type out the question? I don't think you're following it through in the logical order it's intended, but I can't be sure without reading what you're meant to be doing. :smile:

Reply 2

NoCommentMan

I hope it is not due to my language skills (English). It's not my first language.

I typed it as stated in my book.

I understand step 1 and 2 but not step 3. This is when

\frac{1}{(k+1)(k+2)}

comes up.

Reply 3

generalebriety

NoCommentMan

I hope it is not due to my language skills (English). It's not my first language.

I typed it as stated in my book.

I understand step 1 and 2 but not step 3. This is when

\frac{1}{(k+1)(k+2)}

comes up.

No, your English is fine. I'm just saying that it would help to see the question, not just the solution.

(To answer your question, though: just work out

S_{k+1} - S_k

Reply 4

NoCommentMan

generalebriety

No, your English is fine. I'm just saying that it would help to see the question, not just the solution.

(To answer your question, though: just work out

S_{k+1} - S_k

Thanks! I will try that ...

The question was: Proof that the proposition is true for n = 1 and that the proposotion is true for n = k + 1 if it is true for n = k, where k is any positive integer.

Reply 5

Dj-Dark-1

You sure that's the full question? Normally you given a proposition such as 1 + 2 + 3 + ... + m = (m(m+1))/2.

Anyway in step 3, you try to introduce what you assumed in step 2. So in step 3 we have:
(k+1)/(k+2) and we want to introduce k/(k+1) into this. So what do you need to add to k/(k+1) to make (k+1)/(k+2).
We would add 1/((k+1)(k+2))

So in step 3: (k+1)/(k+2) has become Sk + 1/((k+1)(k+2))