Series expansion of xln((x+1)/x)

  • Thread starter damoj
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  • #1
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basically i have to check if
[tex] xln\frac{(x+1)}{x} [/tex]→ 1 as x→∞


the first term is 0 as x→∞

in the answers they say they used maclaurin series and got

[tex]x(\frac{1}{x} + O\frac{1}{1^{2}}) [/tex]

but dont show how they did it.

would the first term in the series be

[tex]a(ln(\frac{a+1}{a}))[/tex]

there a = 0, but then the ln function is defined?

im getting a little confused about have a=0 but we want to figure out where the function goes to as x→∞
 

Answers and Replies

  • #2
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Looks to me it should be:

[tex]1/x(x+O(k))[/tex]

if instead of taking the limit at infinity, we take the limit at zero of the expression:

[tex]1/x\ln\left(\frac{1+1/x}{1/x}\right)[/tex]

Now we can express the log expression as a MacLaurin series.
 
  • #3
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sorry i dont understand what you mean.

are you rewriting our original expression?

how can i write the ln(1+(1/x)) expression so that its defined for x=0
 
  • #4
SammyS
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Why is jackmell's suggestion useful?

[itex]\displaystyle \frac{1+1/x}{1/x}=x+1[/itex]
 
  • #5
Ray Vickson
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sorry i dont understand what you mean.

are you rewriting our original expression?

how can i write the ln(1+(1/x)) expression so that its defined for x=0
You should not care what happens near x = 0; after all, the question is asking about what happens as x --> infinity. So, if x is large, y = 1/x is small, and surely you know how ln(1+y) behaves near y = 0.

RGV
 

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