Simplify e^(1/n+1)/e^(1/n) Homework Equation

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In summary, The conversation is about simplifying the expression e^(1/n+1)/e^(1/n) for a series test "ratio test". The discussion involves combining exponents, finding the LCD, and correcting for a mistake in the subtraction of fractions in the exponent. The ultimate goal is to show that the series in question converges.
  • #1
Jac8897
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Homework Statement



e^(1/n+1)/e^(1/n)


Homework Equations



this is for a series test "ratio test" I need to simplify it I went blank in this part


thanks

The Attempt at a Solution

 
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  • #2
the sum from 1 to infinity I'm guessing?

What are you trying to show?
 
  • #3
i got the answer that this series converges because any number multiply by 0 = 0 but i want to know how to deal with that part look at my attachment

this is the ratio test for calc 2 [An+1/An]
 

Attachments

  • homework.bmp
    96.1 KB · Views: 464
  • #4
Do you know how to simplify

ak/aj by combining their exponents?
 
  • #5
is it a^(k-J)
 
  • #6
So here a=e and what are k and j?
 
  • #7
so e^(1/n+1)-(1/n)
can i do this ?
(1/n+1)-(1/n)

lcd (n+1)(n)
(n^2+n)

so e^(n^2+n)
 
  • #8
I think there are some missing parentheses

By (1/n+1) do you mean 1/(n+1)?

You definitely did not do 1/(n+1)-1/n correctly (or 1/n+1-1/n as order of operations would have it be). Why would you just multiply their denominators?
 
  • #9
sorry I see what i did wrong I attached a picture of what I think would be the correct way can you double check it

thank you
 

Attachments

  • homework1.bmp
    96.1 KB · Views: 443
  • #10
If you post an attachment we have to wait for it to be approved. You might want to put it up at imageshack.us or something and then give us a link to it to make it go faster
 
  • #11
http://img694.imageshack.us/img694/5802/homework1.png

ok thanks for the website
 
Last edited by a moderator:
  • #12
I have to go I will check you replay in one hour thanks for the help
 
  • #13
Check the subtraction of the fractions in the exponent.
 

1. What is the equation for simplifying e^(1/n+1)/e^(1/n)?

The equation for simplifying e^(1/n+1)/e^(1/n) is e^(1/n+1)/e^(1/n) = e^(1/n+1-n) = e^(1/n). This is because when dividing exponential expressions with the same base, the exponents can be subtracted.

2. Can e^(1/n+1)/e^(1/n) be simplified further?

No, e^(1/n+1)/e^(1/n) is already in its simplest form.

3. How can I rewrite e^(1/n+1)/e^(1/n) in a different form?

e^(1/n+1)/e^(1/n) can be rewritten as e^(1/(n+1))/e^(1/n). This is because the exponent rule for adding fractions in the exponent states that e^(a/b+c/b) = e^(a+c)/b.

4. What is the value of e^(1/n+1)/e^(1/n) when n = 0?

The value of e^(1/n+1)/e^(1/n) when n = 0 is undefined. This is because dividing by zero is undefined in mathematics.

5. Can e^(1/n+1)/e^(1/n) be simplified using logarithms?

Yes, e^(1/n+1)/e^(1/n) can be simplified using logarithms. Using the logarithm rule log(a/b) = log(a) - log(b), we can rewrite e^(1/n+1)/e^(1/n) as e^(1/n+1) * e^(-1/n) = e^(1/n+1 - 1/n) = e^(1/n). This is the same result as simplifying using the exponent rule.

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