A telescoping sum is a sum of differences. If an=f(n)−f(n+1), then ∞∑n=1an=(f(1)−f(2))+(f(2)−f(3))+(f(3)−f(4))+…
This adds up to f(1) if limn→∞f(n)=0.
Example: ∞∑n=11n(n+1)=∞∑n=1(1n−1n+1)=11=1,
since limn→∞f(n)=limn→∞1n=0.
Something similar can be said for finite telescoping sums: n∑i=1(f(i)−f(i+1))=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: sn−sn−1=an.
Since derivatives are limits of differences and integrals are limits of sums, we get both versions of the fundamental theorem of calculus: