A telescoping sum is a sum
of differences. We look at a general example. If an=f(n)−f(n+1), then ∞∑n=1an=∞∑n=1(f(n)−f(n+1))=(f(1)−f(2))+(f(2)−f(3))+(f(3)−f(4))+⋯.
DO: Carefully find
the nth partial sum of this series. It is a
telescoping sum. sn=n∑i=1(f(i)−f(i+1))=(f(1)−f(2))+(f(2)−f(3))+⋯+(f(n−1)−f(n))+(f(n)−f(n+1)). You can cancel many
terms (Do this.) to be left
with only the two terms sn=f(1)−f(n+1). Thus we can
compute the convergence/divergence of the series, and if it
converges, we can find its value: ∞∑n=1(f(n)−f(n+1))=limn→∞sn=limn→∞(f(1)−f(n+1))=f(1)−limn→∞f(n+1).
A series with telescoping partial sums is one
of the rare series with which we can compute the value of the
series by using the definition of a series as the limit of its
partial sums. Example: ∞∑n=21n(n−1)=∞∑n=2(1n−1−1n). DO: Check this equality by using partial fraction decomposition on 1n(n−1), then write out
the nth partial sum of this series and cancel terms.
sn=n∑i=2(1i−1−1i)=(11−12)+(12−13)+(13−14)+⋯+(1n−2−1n−1)+(1n−1−1n) =1−1n DO: find the value of the series
by computing the limit of the partial sums.
limn→∞sn=limn→∞(1−1n)=1−0=1.
So ∞∑n=21n(n−1)=1.
.
This video will play with these ideas in a more abstract way, and
will tie them to integrals, derivatives, and the FTC.
Depending upon your instructor, the information in the video, while
interesting, is optional.