Difference between revisions of "Telescoping series"

Latest revision as of 15:37, 21 July 2024

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:

$\sum_{k=1}^{n}k=\sum_{k=1}^{n}(a_k-a_{k-1})=a_n-a_0=\frac{n(n+1)}{2}-\frac{0\cdot1}{2}=\boxed{\frac{n(n+1)}{2}}$ .

(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:

$\sum_{k=1}^{n}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{n}(a_k-a_{k+1})=a_1-a_{n+1}=1-\frac{1}{n+1}=\boxed{\frac{n}{n+1}}$ .

Problems

Introductory

When simplified the product $\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)$ becomes:

$\textbf{(A)}\ \frac1n \qquad\textbf{(B)}\ \frac2n\qquad\textbf{(C)}\ \frac{2(n-1)}{n}\qquad\textbf{(D)}\ \frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)}$ (Source)

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!)

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

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

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

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

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

(Source)

Intermediate

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)

Olympiad

Find the value of $\sum_{k=2}^{\infty} \left( \zeta(k) - 1 \right)$ , where $\zeta$ is the Riemann zeta function $\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.$

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-In mathematics, a telescoping series is a series whose partial sums eventually only have a finite number of terms after cancellation. For example, let's try to find value of the series <math>\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + ... + \frac{1}{(n-1) \cdot n}</math>. We can see that <math>\frac{1}{n-1} - \frac{1}{n} = \frac{1}{(n-1) \cdot n}</math>. Thus, <math>\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + ... + \frac{1}{(n-1) \cdot n}</math> = <math>\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{n-1} - \frac{1}{n}</math>. Then, we can see that all of the terms except <math>\frac{1}{1}</math> and <math>\frac{1}{n}</math>. So the answer is <math>1 - \frac{1}{n} = \frac{n-1}{n}</math>. We can see that in the process, we manipulated a large series so the many terms cancelled out with each other, leaving only a few terms that we could easily calculate with. This is usually how most telescoping series work.
+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 <math>\sum_{k=1}^{n} f(k)=\sum_{k=1}^{n}(a_{k}-a_{k-1})=a_{k}-a_0</math> for some expression <math>a_{k}</math>.
+==Example 1==
+Derive the formula for the sum of the first <math>n</math> counting numbers.
+==Solution 1==
+We wish to write <math>\sum_{k=1}^{n}k=\sum_{k=1}^{n}(a_k-a_{k-1})</math> for some expression <math>a_k</math>. This expression is <math>a_k=\frac{k(k+1)}{2}</math> as <math>a_k-a_{k-1}=k</math>.
+We then telescope the expression:
+<math>\sum_{k=1}^{n}k=\sum_{k=1}^{n}(a_k-a_{k-1})=a_n-a_0=\frac{n(n+1)}{2}-\frac{0\cdot1}{2}=\boxed{\frac{n(n+1)}{2}}</math>.
+(Notice how the sum telescopes— <math>\sum_{k=1}^{n}(a_k-a_{k-1})</math> contains a positive and a negative of every value of <math>a_k</math> from <math>k=1</math> to <math>k=n-1</math>, so those terms cancel. We are then left with <math>a_n-a_0</math>, the only terms which did not cancel.)
+==Example 2==
+Find a general formula for <math>\sum_{k=1}^{n}\left(\frac{1}{k(k+1)}\right)</math>, where <math>n\in \mathbb{Z}^{+}</math>.
+==Solution 2==
+We wish to write <math>\sum_{k=1}^{n}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{n}(a_k-a_{k+1})</math> for some expression <math>a_k</math>. This can be easily achieved with <math>a_k=\frac{1}{k}</math> as <math>a_k-a_{k+1}=\frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}</math> by simple computation.
+We then telescope the expression:
+<math>\sum_{k=1}^{n}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{n}(a_k-a_{k+1})=a_1-a_{n+1}=1-\frac{1}{n+1}=\boxed{\frac{n}{n+1}}</math>.
 ==Problems==
+===Introductory===
+*When simplified the product <math>\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)\cdots\left(1-\frac1n\right)</math> becomes:
+<math>\textbf{(A)}\ \frac1n \qquad\textbf{(B)}\ \frac2n\qquad\textbf{(C)}\ \frac{2(n-1)}{n}\qquad\textbf{(D)}\ \frac{2}{n(n+1)}\qquad\textbf{(E)}\ \frac{3}{n(n+1)} </math>  ([[1959 AHSME Problems/Problem 37|Source]])
+*The sum <math>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}</math> can be expressed as <math>a-\frac{1}{b!}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>? ([[2022 AMC 10B Problems/Problem 9|Source]])
+*Which of the following is equivalent to <math>(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})?</math> (Hint: difference of squares!)
+<math>\textbf{(A)} ~3^{127} + 2^{127}</math>
+<math>\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}</math>
+<math>\textbf{(C)} ~3^{128}-2^{128}</math>
+<math>\textbf{(D)} ~3^{128} + 2^{128}</math>
+<math>\textbf{(E)} ~5^{127}</math>
+([[2021 AMC 12A Problems/Problem 9|Source]])
 ===Intermediate===
-*Find the value of <math>\sum_{k=2}^{\infty} \left( \zeta(k) - 1 \right),</math> where <math>\zeta</math> is the [[Riemann zeta function]] <math>\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.</math>
+*Let <math>S</math> denote the value of the sum <math>\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}.</math> <math>S</math> can be expressed as <math>p + q \sqrt{r}</math>, where <math>p, q,</math> and <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime. Determine <math>p + q + r</math>. ([[Mock AIME 3 Pre 2005 Problems/Problem 6|Source]])
+===Olympiad===
+*Find the value of <math>\sum_{k=2}^{\infty} \left( \zeta(k) - 1 \right)</math>, where <math>\zeta</math> is the [[Riemann zeta function]] <math>\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}.</math>
+==See Also==
+* [[Algebra]]
+* [[Summation]]
+[[Category:Algebra]]

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