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

basically i have to check if
$$xln\frac{(x+1)}{x}$$→ 1 as x→∞

the first term is 0 as x→∞

in the answers they say they used maclaurin series and got

$$x(\frac{1}{x} + O\frac{1}{1^{2}})$$

but dont show how they did it.

would the first term in the series be

$$a(ln(\frac{a+1}{a}))$$

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→∞

Looks to me it should be:

$$1/x(x+O(k))$$

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

$$1/x\ln\left(\frac{1+1/x}{1/x}\right)$$

Now we can express the log expression as a MacLaurin series.

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

SammyS
Staff Emeritus
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Why is jackmell's suggestion useful?

$\displaystyle \frac{1+1/x}{1/x}=x+1$

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