A telescoping sum is a sum of differences. If $a_n = f(n)-f(n+1)$, then $$\sum_{n=1}^\infty a_n = \left(f(1)-f(2)\right)+\left(f(2)-f(3)\right)+\left(f(3)-f(4)\right) + \ldots$$This adds up to $f(1)$ if $\displaystyle\lim_{n\to\infty}f(n) =0$.
Something similar can be said for finite telescoping sums: $$\sum_{i=1}^n \left(f(i)-f(i+1)\right) = f(1)-f(n+1).$$ That is, the sum of the differences is the change in the original function.
Likewise, the difference of successive partial sums is the original function: $s_{n}-s_{n-1}=a_n$.
Since derivatives are limits of differences and integrals are limits of sums, we get both versions of the fundamental theorem of calculus:
$$\int_a^b f'(x)\, dx = \hbox{limit of sum of differences} = f(b)-f(a);$$ $$\frac{d}{dx}\left(\int_1^x f(t)\, dt\right) = \hbox{limit of difference of sums} = f(x)$$