Help with induction proof 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.
 A: 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
$$\frac{n}{2n+1}+\frac{1}{(2n+1)(2n+3)}\stackrel{?}=\frac{n+1}{2(n+1)+1}$$
A: The second summand in your last line must be $$\frac{1}{(2(k+1)-1)(2(k+1)+1)}$$
A: 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.
A: $$
\\\frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=
\\\frac{1}{2}\cdot((\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\cdots+(\frac{1}{2n-1}-\frac{1}{2n+1}))=
\\\frac{1}{2}\cdot(1-\frac{1}{2n+1})=\frac{n}{2n+1}
$$
