Telescoping series

In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. This is often done by using a form of $\sum_{k=1}^{n} f(k)=\sum_{k=1}^{n}(a_{k}-a_{k-1})=a_{k}-a_0$ for some expression $a_{k}$.

Example 1

Derive the formula for the sum of the first $n$ counting numbers.

Solution 1

We wish to write $\sum_{k=1}^{n}k=\sum_{k=1}^{n}(a_k-a_{k-1})$ for some expression $a_k$. This expression is $a_k=\frac{k(k+1)}{2}$ as $a_k-a_{k-1}=k$.

We then telescope the expression:


(Notice how the sum telescopes— $\sum_{k=1}^{n}(a_k-a_{k-1})$ contains a positive and a negative of every value of $a_k$ from $k=1$ to $k=n-1$, so those terms cancel. We are then left with $a_n-a_0$, the only terms which did not cancel.)

Example 2

Find a general formula for $\sum_{k=1}^{n}\left(\frac{1}{k(k+1)}\right)$, where $n\in \mathbb{Z}^{+}$.

Solution 2

We wish to write $\sum_{k=1}^{n}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{n}(a_k-a_{k+1})$ for some expression $a_k$. This can be easily achieved with $a_k=\frac{1}{k}$ as $a_k-a_{k+1}=\frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}$ by simple computation.

We then telescope the expression:




  • The sum $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}$ can be expressed as $a-\frac{1}{b!}$, where $a$ and $b$ are positive integers. What is $a+b$? (Source)

  • Which of the following is equivalent to $(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?$ (Hint: difference of squares!)

$\textbf{(A)} ~3^{127} + 2^{127}$

$\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}$

$\textbf{(C)} ~3^{128}-2^{128}$

$\textbf{(D)} ~3^{128} + 2^{128}$

$\textbf{(E)} ~5^{127}$



  • Let $S$ denote the value of the sum $\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}.$ $S$ can be expressed as $p + q \sqrt{r}$, where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime. Determine $p + q + r$. (Source)


See Also