Difference between revisions of "Telescoping series"
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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. | + | |
| + | 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>. | ||
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| + | ==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>. | ||
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| + | We then telescope the expression: | ||
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| + | <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>. | ||
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| + | (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>. | ||
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| + | ==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. | ||
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| + | We then telescope the expression: | ||
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| + | <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>. | ||
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| + | |||
| + | ==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]]) | ||
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| + | *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]]) | ||
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| + | *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!) | ||
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| + | <math>\textbf{(A)} ~3^{127} + 2^{127}</math> | ||
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| + | <math>\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63}</math> | ||
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| + | <math>\textbf{(C)} ~3^{128}-2^{128}</math> | ||
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| + | <math>\textbf{(D)} ~3^{128} + 2^{128}</math> | ||
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| + | <math>\textbf{(E)} ~5^{127}</math> | ||
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| + | ([[2021 AMC 12A Problems/Problem 9|Source]]) | ||
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| + | ===Intermediate=== | ||
| + | *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]]) | ||
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| + | |||
| + | ===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> | ||
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| + | |||
| + | ==See Also== | ||
| + | * [[Algebra]] | ||
| + | * [[Summation]] | ||
| + | |||
| + | [[Category:Algebra]] | ||
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 
for some expression 
.
Contents
Example 1
Derive the formula for the sum of the first 
counting numbers.
Solution 1
We wish to write 
for some expression 
. This expression is 
as 
.
We then telescope the expression:
.
(Notice how the sum telescopes— 
contains a positive and a negative of every value of from 
to 
, so those terms cancel. We are then left with 
, the only terms which did not cancel.)
Example 2
Find a general formula for 
, where 
.
Solution 2
We wish to write 
for some expression . This can be easily achieved with 
as 
by simple computation.
We then telescope the expression:
.
Problems
Introductory
- When simplified the product
becomes:
becomes:
(Source)
- The sum
can be expressed as 
, where 
and 
are positive integers. What is 
? (Source)
can be expressed as 
, where 
and 
are positive integers. What is 
? (
- Which of the following is equivalent to
(Hint: difference of squares!)
(Hint: difference of squares!)
(Source)
Intermediate
- Let
denote the value of the sum 
can be expressed as 
, where 
and 
are positive integers and is not divisible by the square of any prime. Determine 
. (Source)
denote the value of the sum 
, where 
and 
are positive integers and 
. (
Olympiad
- Find the value of
, where 
is the Riemann zeta function 
, where 
is the 
