Limit ln(n-1/n+1) as n->infinity

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Homework Statement


Hi! I need help confirming the limit of ln(n-1/n+1) as n->infinity.

If you multiply top and bottom of the quotient by 1/n you'd end up with ln(1) = 0, no? I must be missing something rather simple here because my hp50 won't even compute. Thanks!

Homework Statement

 
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AFinch said:

Homework Statement


Hi! I need help confirming the limit of ln(n-1/n+1) as n->infinity.

If you multiply top and bottom of the quotient by 1/n you'd end up with ln(1) = 0, no? I must be missing something rather simple here because my hp50 won't even compute. Thanks!

Homework Statement


Yes, you did it correctly. I'm not sure why the hp50 has problems.
 
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Thank you! After getting your response I looked further into the problem with the hp and figured it out. I had a 1. instead of 1 (sans decimal), and it won't take a limit with the decimal because it's a "real" number. This was of much help.
 
AFinch said:

Homework Statement


Hi! I need help confirming the limit of ln(n-1/n+1) as n->infinity.

If you multiply top and bottom of the quotient by 1/n you'd end up with ln(1) = 0, no? I must be missing something rather simple here because my hp50 won't even compute. Thanks!

Homework Statement


You should realize that what you have written is ln[n + 1 - 1/n], which has no limit. Did you really mean ln[(n-1)/(n+1)]? If so, use brackets!

RGV
 
Yes ln[(x-1)/(x+1)] is what I actually meant, thanks for correcting my mistake.
 
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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