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

damoj
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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 don't 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→∞
 
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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 don't 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
 
Why is jackmell's suggestion useful?

\displaystyle \frac{1+1/x}{1/x}=x+1
 
damoj said:
sorry i don't 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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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