Prove that for any positive integer n the following identity holds, 
am i supposed to use mathematical induction to prove this?
If i do, i have no idea how to start the proof.
any tips or suggestions would be great!
Thank you!
 A: Hint: Add $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}$ to both sides, and check that you get the same result.
When you are adding on the right, note that you are adding two copies of $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}$.
Remark: One can also do an induction proof. When we increment $n$ by $1$, the left-hand side increases by an amount $\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$. and the right-hand side increases by an amount $\frac{1}{2n+1}-\frac{1}{2n+2}$. These two quantities are easily verified to be equal.
A: Since the following 2 expressions
$$\begin{align}
\text{LHS}_{n+1} - \text{LHS}_{n} = & \frac{1}{2n+2} + \frac{1}{2n+1} - \frac{1}{n+1}\\
\text{RHS}_{n+1} - \text{RHS}_{n} = & \frac{1}{2n+1} - \frac{1}{2n+2}
\end{align}$$
clearly equal to each other, we have
$$\text{LHS}_{n+1} - \text{RHS}_{n+1} = \text{LHS}_{n} - \text{RHS}_{n}$$
Since $\text{LHS}_1 = \frac{1}{1+1} = 1 - \frac{1}{2} = \text{RHS}_1$, we have $\text{LHS}_n = \text{RHS}_n$ for all $n$.
