Can't find formula of a serie.

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Homework Help Overview

The discussion revolves around finding the formula for a series given its first five terms, specifically in the context of calculus. The series presented is 1 + (2/5)² + (3/8)³ + (4/11)⁴ + (5/14)⁵ + ... The original poster expresses uncertainty about the correct formula, particularly due to the first term being 1 instead of a suggested 1/2.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore various potential formulas for the series and question whether the first term might have been misprinted. Some suggest alternative series representations and discuss the implications of adjusting the first term on convergence.

Discussion Status

Participants are actively engaging with the problem, with some offering potential formulations and others reflecting on the implications of convergence related to the series. There is no explicit consensus on the correct formula, but productive lines of reasoning are being explored.

Contextual Notes

There is a mention of a theorem regarding the convergence of series when adding or subtracting finite terms, which is relevant to the discussion of the series in question. The original poster also notes that the answer is not provided in the book, adding to the uncertainty.

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A problem from my old Calculus book I can't solve...

Homework Statement



Find the formula of this serie knowing its first five terms:

[tex]1 +\left(\frac{2}{5}\right)^{2} +\left(\frac{3}{8}\right)^{3} +\left(\frac{4}{11}\right)^{4} +\left(\frac{5}{14}\right)^{5} + \cdots[/tex]

2. Relevant formulas

If the first term was 1/2 then the formula would simply be

[tex]\sum_{i = 1}^{\infty}\left(\frac{i}{2+3(i-1)}\right)^{i}[/tex]

but the first term being 1, I can't see how to do it...

(Maybe the first term IS 1/2 and it got missprinted in the book ?)

Also, the answer is not given in the book.

The Attempt at a Solution



I've tried these as possible values for a first item equal to 1 without any success:

[tex]\left(\frac{1}{2}\right)^{0}=1[/tex]

[tex]\left(\frac{1}{1}\right)^{1}=1[/tex]

[tex]\left(\frac{2}{2}\right)^{1}=1[/tex]


So, is there a formula for this serie with first term = 1 ?

Thanks !
 
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could you do 1 + [tex] \sum_{i = 2}^{\infty}\left(\frac{i}{3i-1)}\right)^{i}[/tex] ? I can't think of a series that fits there without just not counting the "first" term.
 
How about this?
[tex]1/2 +\left(\frac{1}{2}\right)^{1} + \left(\frac{2}{5}\right)^{2} +\left(\frac{3}{8}\right)^{3} +\left(\frac{4}{11}\right)^{4} +\left(\frac{5}{14}\right)^{5} + \cdots[/tex]

Now you can write this as 1/2 + the summation you had, starting at i = 1, though. I don't see any problem with doing that.
 
Thanks Hogger and Mark44.

I think you got it !

I needed the formula of the serie to figure out if it converge or not.
Substracting 1/2 from it made me remember about this theorem from the book:

"Convergence or divergence of a serie is not affected by adding or substracting a finite number of terms from it."

So, and if I get all this right, convergence or divergence of these series will be the same:

The serie from the book:

[tex]1 +\left(\frac{2}{5}\right)^{2} +\left(\frac{3}{8}\right)^{3} +\left(\frac{4}{11}\right)^{4} +\left(\frac{5}{14}\right)^{5} + \cdots[/tex]

The serie minus 1/2:

[tex]\left(\frac{1}{2}\right)^{1} +\left(\frac{2}{5}\right)^{2} +\left(\frac{3}{8}\right)^{3} +\left(\frac{4}{11}\right)^{4} +\left(\frac{5}{14}\right)^{5} + \cdots[/tex]

The second serie brings us back to the Relevant formula of my original post and I know how to find-out if it converge or not.

If it does, the serie from the book does to.

Is this right ?

Thanks !
 
Yes, this is right.

BTW, there is no such word in English as "serie." The word is series, which is both singular and plural.
 

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