One interesting method that can be used to evaluate infinite sums $$\sum_{x=1}^\infty f(x)$$ is to find a function $g(x)$ that has the property $$g(x)-g(x+1)=f(x)$$ because if we substitute this into the infinite sum, it telescopes and we are left with just $g(1)$ as the value of our sum.
For example, consider the sum $$\sum_{x=1}^\infty \frac{1}{2^x}=1$$ if we let $$f(x)=\frac{1}{2^x}$$ then we have $$g(x+1)=g(x)-\frac{1}{2^x}; g(1)=1$$ and we can conclude that $$g(x)=\frac{1}{2^{x-1}}$$ so the value of the sum is $g(1)$, which is $1$.
Can anybody think of any other ways to "split up" functions that are less easy to split than exponential ones? Like polynomials or rational functions?
Such a series is called a telescoping series.
For example $$\sum_{i=1}^n i(i+1)=\frac{n(n+1)(n+2)}3$$, which may be written $$\sum_{i=1}^n {}_iP_2=\dfrac{{}_{n+1}P_3}{3}$$ or $$\sum_{i=1}^n \binom{i}2=\binom{n+1}{3}$$. This generalises: $$\sum_{i=1}^n {}_iP_r=\dfrac{{}_{n+1}P_{r+1}}{r+1}$$ or $$\sum_{i=1}^n \binom{i}{r}=\binom{n+1}{r+1}$$