Finding the Sum of a Series: e^{(n+1)/n}- e^{(n+2)/(n+1)}

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In summary: It sounds like you might need to find the limits of the series. Have you considered the possibility that it's a cheap trick? Possibly a telescoping series?Very close!But, alas, putting numbers in for n for the first three terms gives...(2 - 1.5) + (1.125 - .88889) + (.39506 - .3255208)The powers are throwing it off...Uh, n=1. e^(2)-e^(3/2). n=2. e^(3/2)-e^(4/3). n=3. e^(4/3)-e^(5/4
  • #1
the7joker7
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Homework Statement



Find the sum of the series e[tex]{(n + 1)/n)}[/tex] - e[tex]^{(n + 2)/(n + 1)}[/tex]

The Attempt at a Solution



Well for starters I got it into the proper format...IE

(((n + 1)/(n))^n)/n! - (((n + 2)/(n + 1))^n)/n!

But then I get a little lost...I would know how to take the limit, find convergence, divergence, but finding sums can be hard...I mean, it doesn't converge fast enough to just take the first few partial sums and be satisfied with that answer...help?
 
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  • #2
Have you considered the possibility that it's a cheap trick? Possibly a telescoping series?
 
  • #3
Very close!

But, alas, putting numbers in for n for the first three terms gives...

(2 - 1.5) + (1.125 - .88889) + (.39506 - .3255208)

The powers are throwing it off...
 
  • #4
Uh, n=1. e^(2)-e^(3/2). n=2. e^(3/2)-e^(4/3). n=3. e^(4/3)-e^(5/4). What do you mean 'the powers are throwing it off'? You are the one who is 'throwing it off'.
 
  • #5
In the series expansion e^((n+1)/n)=sum(((n+1)/n)^k/k! over k). k and n aren't the same thing.
 
  • #6
Perhaps you would see what Dick is suggesting if you rewrote the summand as [tex]e ( e^{\frac{1}{n}} - e^{\frac{1}{n+1}})[/tex].
 
  • #7
Ohhh...I see.

So, what is k supposed to be? Is it just a variable, or does it get a value?
 
  • #8
k is summed over in each exponential. They are both infinite sums. You've only selected the nth term from each sum. There are many more terms.
 
  • #9
Lemme see here...

The series is indeed telescoping and I have it simplifying to...

((2^k)/k!) - (((n + 2)/(n + 1))^k)/k!

Since n + 2 over n + 1 will run to 1...

((2^k/k!) - ((1^k/k!))

So...

1^k/k!

Am I on the right track?
 
  • #10
No! That's not 'simplifying' it at all, except where it's wrong. You don't need to series expand the exponentials at all! The original series telescopes!
 
  • #11
the7joker7 said:

Homework Statement



Find the sum of the series e[tex]{(n + 1)/n)}[/tex] - e[tex]^{(n + 2)/(n + 1)}[/tex]
There is a missing parenthesis here. Or do you mean
[tex]e^{(n+1)/n}- e^{(n+2)/(n+1)}[/tex]?

The Attempt at a Solution



Well for starters I got it into the proper format...IE

(((n + 1)/(n))^n)/n! - (((n + 2)/(n + 1))^n)/n!
What do you mean by "proper format"?

But then I get a little lost...I would know how to take the limit, find convergence, divergence, but finding sums can be hard...I mean, it doesn't converge fast enough to just take the first few partial sums and be satisfied with that answer...help?
Actually write out the first few terms. An think about what Dick said.
 
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Related to Finding the Sum of a Series: e^{(n+1)/n}- e^{(n+2)/(n+1)}

What is the definition of a series?

A series is a sequence of numbers or terms that are added together in a specific order.

What is the formula for finding the sum of a series?

The formula for finding the sum of a series is S = a1 + a2 + a3 + ... + an, where S is the sum, a1 is the first term, a2 is the second term, and so on until an is the last term in the series.

What is the difference between an arithmetic and geometric series?

An arithmetic series has a constant difference between each term, while a geometric series has a constant ratio between each term.

How do you determine if a series is convergent or divergent?

A series is convergent if the sum of all the terms approaches a finite number as the number of terms increases. A series is divergent if the sum of the terms does not approach a finite number.

What are some common techniques for finding the sum of a series?

Some common techniques for finding the sum of a series include using the formula for arithmetic or geometric series, using the partial sum formula, or using the ratio or root test to determine convergence or divergence.

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