Can someone help me on this series?

  • Thread starter egwin
  • Start date
  • #1
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Hello,
I want to check convergence for the following series:
∑2^(1/n)-1
so that's the n'th root of two minus one, for n going to positive infinity

Now all possible formulas i know come out inconclusive,
so the only formula i can think of is the integer formula, now i thought
to make things easier to just integrate 2^(1/n) and if this series proves to be convergent then that proves the convergency (as it is always "1" larger than the other one).

Now as i am seem to be a complete idiot i cannot even find the integer for this one.

Can someone help me?

Sorry if my English is not so good.

Thank you in advance,
Egwin.
 

Answers and Replies

  • #2
StatusX
Homework Helper
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You have the expansion:

[tex] (1+x)^p = 1 +p x + O(p^2) [/tex]

What happens when x=1 and p=1/n is very small?
 
  • #3
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thanks

i didnt really know what you meant,

but i have the solution now,
if you're interested, it's in the attachment.

egwin.
 

Attachments

  • #4
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hello status x,

in your expansion /x/ should be <1
does it still work for x=1 then?
 
  • #5
StatusX
Homework Helper
2,564
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The full expansion is:

[tex](1+x)^p=1+px+\frac{p(p-1)}{2!} x^2 + \frac{p(p-1)(p-2)}{3!} x^3 + ...[/tex]

which is valid for all p, all |x|<1. I made a mistake before, the second term is actually still O(p) as p->0. For p very small, we can approximate [itex]p-n \approx -n[/itex], so:

[tex](1+x)^p \approx 1+px(1+\frac{-1}{2} x + \frac{2\cdot 1}{3!} x^2 + ...)[/tex]

[tex] =1+px(1-\frac{1}{2} x + \frac{1}{3} x^2 + ...)[/tex]

with x=1 (or, more accurately, taking the limit as x->1, since, as you pointed out, the series doesn't strictly converge at x=1), the term in parantheses converges to some number (ln(2) I think), so we still get:

[tex]2^p \rightarrow 1+Cp[/tex] as [itex]p \rightarrow 0[/itex]

for some constant C, or in other words:

[tex]2^{1/n}-1 \rightarrow \frac{C}{n}[/tex] as [itex]n \rightarrow \infty[/itex]

Since the sum of 1/n diverges, this implies the sum of 2^(1/n)-1 does as well. I think this is all right, but there are some iffy arguments, and the method you posted is probably easier.
 
Last edited:

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