Does anyone know an infinite series summation that is to 1/5 or 1/7?

In summary, there are various methods for finding infinite series summations that are equal to 1/5 or 1/7, including using geometric series, trigonometric series, and normalizing factors. There are also methods for creating a series that can cover all whole integer fractions by changing the initial value, index of summation, or lower limit of summation. These methods can be found in resources such as comprehensive tables of integrals and series expansions.
  • #1
mesa
Gold Member
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The title pretty much says it all, does anyone know infinite series summations that are equal to 1/5 or 1/7?
 
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  • #2
You can easily invent a geometric series that sums to any value you want.

Or you could do something like the series expansion of ##\frac 1 5 \sin \pi##.
 
  • #3
AlephZero said:
You can easily invent a geometric series that sums to any value you want.

Or you could do something like the series expansion of ##\frac 1 5 \sin \pi##.

Excellent! We can get those by changing the initial value and inputting different integers into the series.

Do you know of any other ways to get to these values outside of geometric and trigonometric series? For example do we have a factorial series for calculating these values?
 
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  • #4
Any function whose value you can calculate and whose Taylor series can be expanded will give you a series. What is the purpose of this question?
 
  • #5
AlephZero said:
Or you could do something like the series expansion of ##\frac 1 5 \sin \pi##.
That won't get him 1/5, though. :tongue:

I'm sure that was a typo. mesa, he probably meant ##\frac{\sin\frac{\pi}{2}}{5}##.
 
  • #6
How about the trivial example where the first term in the series is 1/5 and all the remaining terms are zero? No need to get all fancy here guys!
 
  • #7
jgens said:
How about the trivial example where the first term in the series is 1/5 and all the remaining terms are zero? No need to get all fancy here guys!
Indeed.

mesa, you should look back at your other thread on infinite series. Many of the same series (plural) apply here.
 
  • #8
Get any sum that yields any finite number, multiply it by a "normalizing factor".

[tex]e=\sum_0^\infty\frac 1 {n!}[/tex]

so

[tex]\frac 1 5 = \sum_0^\infty\frac 1 {5en!}[/tex]
 
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  • #10
Office_Shredder said:
Any function whose value you can calculate and whose Taylor series can be expanded will give you a series. What is the purpose of this question?

To get an idea of what is out there. Do we have a 'general solution' outside of geometric series, trig, and normalizing factors for solving all whole integer fractions with '1' in the numerator through infinite series?

Mandelbroth said:
That won't get him 1/5, though. :tongue:

I'm sure that was a typo. mesa, he probably meant ##\frac{\sin\frac{\pi}{2}}{5}##.

Very good, it can be tough to spot errors late on New Years Eve... :)

jgens said:
How about the trivial example where the first term in the series is 1/5 and all the remaining terms are zero? No need to get all fancy here guys!

Hah!

Mandelbroth said:
Indeed.

mesa, you should look back at your other thread on infinite series. Many of the same series (plural) apply here.

Yes they do, I was curious if we had more 'general' solutions for infinite series that would give whole integer fractions.

Borek said:
Get any sum that yields any finite number, multiply it by a "normalizing factor".

[tex]e=\sum_0^\infty\frac 1 {n!}[/tex]

so

[tex]\frac 1 5 = \sum_0^\infty\frac 1 {5en!}[/tex]

That looks familiar :)

Seydlitz said:
One mentor gave me this link a couple of months ago, it's basically a comprehensive table of integrals and sum of series. You can find pretty interesting series there.

http://atsol.fis.ucv.cl/dariop/sites...ij_Engl._2.pdf

Hmmm, that link doesn't seem to be working?
 
  • #11
Mandelbroth said:
I'm sure that was a typo.

More of a brain-fart than a typo. I divided ##2\pi## by ##2## and got ##\pi##. Full marks for arithmetic, but zero marks for doing the wrong sum. :redface:
 
  • #12
1/5 = 0.19999...

1/7 = 0.142857142857...
 
  • #13
AlephZero said:
More of a brain-fart than a typo. I divided ##2\pi## by ##2## and got ##\pi##. Full marks for arithmetic, but zero marks for doing the wrong sum. :redface:
Nah. We all make mistakes or have lapses in genius from time to time. It's all good. :biggrin:

mesa said:
Yes they do, I was curious if we had more 'general' solutions for infinite series that would give whole integer fractions.
What do you mean "more general"?
 
  • #14
Mandelbroth said:
What do you mean "more general"?

I should probably have said, 'general solution' for an infinite series that can be used to get any whole integer fraction by changing the index of summation. For example, an infinite series that would be able to compute 1/5 and 1/7 simply by changing the index. Do you know if we have anything like this?
 
  • #15
If you want a series that converges to c, take a series that converges to b, and add this term: (c-b) to the front.
 
  • #16
Robert1986 said:
If you want a series that converges to c, take a series that converges to b, and add this term: (c-b) to the front.

That certainly would work much like Boreks suggestion of using a 'normalizing factor'.

Do you know of any series that will equal all whole integer fractions by simply changing the lower limit of summation?
 
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  • #17
mesa said:
That certainly would work much like Boreks suggestion of using a 'normalizing factor'.

Do you know of any series that will equal all whole integer fractions by simply changing the lower limit of summation?

Consider the sequence $$a_i=\{1,-1,\frac{1}{2},\frac{1}{2},-1,\frac{1}{3},\frac{1}{3},\frac{1}{3},-1,...\}.$$

##\displaystyle \sum_{n_0\leq i\leq n}a_i## should cover everything you want. :tongue:
 
  • #18
Mandelbroth said:
Consider the sequence $$a_i=\{1,-1,\frac{1}{2},\frac{1}{2},-1,\frac{1}{3},\frac{1}{3},\frac{1}{3},-1,...\}.$$

##\displaystyle \sum_{n_0\leq i\leq n}a_i## should cover everything you want. :tongue:

Brilliant!
...and I need to brush up on my writing comprehension skills, you guys are too clever :redface:
 

1. What is an infinite series summation?

An infinite series summation is a mathematical expression that represents the sum of an infinite number of terms. It is written in the form of ∑n=1 an, where an is the nth term of the series.

2. Can an infinite series summation be equal to 1/5 or 1/7?

Yes, an infinite series summation can be equal to 1/5 or 1/7. It depends on the specific series and the values of its terms.

3. What is the significance of an infinite series summation being equal to 1/5 or 1/7?

An infinite series summation being equal to 1/5 or 1/7 can have different mathematical implications depending on the context. It could represent a specific value or convergence of a series, which can be used in various mathematical calculations and formulas.

4. How do you find an infinite series summation that is equal to 1/5 or 1/7?

There is no one specific method for finding an infinite series summation that is equal to 1/5 or 1/7. It requires knowledge of mathematical concepts and techniques such as convergence tests, geometric series, and telescoping series.

5. Can you give an example of an infinite series summation that is equal to 1/5 or 1/7?

One example of an infinite series summation that is equal to 1/5 is the series ∑n=1 1/(5n). This series is a geometric series with a common ratio of 1/5, and it converges to 1/4. An example of an infinite series summation that is equal to 1/7 is the series ∑n=1 1/(7n). This series is also a geometric series with a common ratio of 1/7, and it converges to 1/6.

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