Help with induction proof

Question

I need help with the following induction proof which I am not sure if I am doing correctly.

$$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$$

I check for $n=1$ (Base case)

$$\frac{1}{1\cdot3}=\frac13$$

$$\frac{1}{2\cdot1+1}=\frac13$$

Now, is this the correct next step in my proof?

$$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1)-1)(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$

I we assume it is correct for $n=k$ then it is also true for $n=k+1$ which means that the RHS must be equal to the LHS.

Answer 1

For the induction step we need to prove that

$$\frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1} \\\implies \frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n+1)(2n+3)}=\frac{n+1}{2(n+1)+1}$$

then we have

$$\frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n+1)(2n+3)}\stackrel{Ind. Hyp.}=\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\stackrel{?}=\frac{n+1}{2(n+1)+1}$$

then all reduces to prove that

$$\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\stackrel{?}=\frac{n+1}{2(n+1)+1}$$

Answer 2

The second summand in your last line must be $$\frac{1}{(2(k+1)-1)(2(k+1)+1)}$$

Answer 3

Your idea is correct but $$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1))(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$

Should have been $$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1)-1)(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$

Good luck finishing it up.

Answer 4

$$ \\\frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n-1)(2n+1)}= \\\frac{1}{2}\cdot((\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\cdots+(\frac{1}{2n-1}-\frac{1}{2n+1}))= \\\frac{1}{2}\cdot(1-\frac{1}{2n+1})=\frac{n}{2n+1} $$