Proof by induction: $\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)$

Hint $ $ For $ \, f(n)=\rm RHS - LHS\ $ show $ \ f(n\!+\!1) - f(n) = 0 = f(1)$ $\overset{\rm induct}\Longrightarrow\ f(n) = f(1) = 0\,.$

Note $ \ f(n\!+\!1)\ =\ f(n)\ $ boils down to $ \displaystyle\ \frac{2}{2n\!+\!2} = \frac{1}{n\!+\!1}\ $ by a very simple algebraic calculation.

Note how a simple transformation reduces inductive proof to the triviality that a function on $ \,\mathbb N\,$ is constant iff its first difference vanishes, i.e. $ \ f(n\!+\!1)-f(n)\ =\ 0.\,$ The inductive proof of this boils down to equality transitivity on a telescoping chain of inequalities $\,f(1) = f(2) = \,\cdots\, = f(n).\,$ While it is intuitively "obvious" that a function that never changes $ \,f(n\!+\!1) = f(n)\,$ is constant $ \,f(n) = f(1)\,$ for all $\,n\in\Bbb N\,$ it does require rigorous proof!

Many inductive proofs aimplify when handled similarly - by transforming the problem so that the induction becomes completely trivial, e.g. that a product of terms $> 1$ is $> 1$. Often, as above, the reduced problem succumbs to a trivial telescopic proof. You can find $\,$ many examples of telescopy in my prior posts here. These methods often serve to reduce inductions to problems that have purely mechanical solution - solvable by using rote algebra with no ingenuity required (e.g. by algorithms in computer algebra systems). For example, the above problem reduces to verifying that a rational function is zero, which requires only simple mechanical polynomial arithmetic.