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. 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. |
==Problems== | ==Problems== | ||
===Intermediate=== | ===Intermediate=== | ||
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*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> | *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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Revision as of 19:35, 10 September 2023
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 
. We can see that 
. Thus, = 
. Then, we can see that all of the terms except 
and 
. So the answer is 
. 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.
Problems
Intermediate
- Find the value of
where 
is the Riemann zeta function 
where 
is the 
