Can't find formula of a serie.

  • #1
bundleguide
4
0
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 !
 

Answers and Replies

  • #2
Hogger
21
0
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.
 
  • #3
36,216
8,199
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.
 
  • #4
bundleguide
4
0
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 !
 
  • #5
36,216
8,199
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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