Sum of Series with Telescoping Equation

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Homework Help Overview

The problem involves finding the sum of an infinite series defined by the expression \(\sum^{\infty}_{0}(e^{\frac{n+1}{n}}-e^{\frac{n+2}{n+1}})\), which is noted to relate to telescoping series.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the nature of the series as telescoping and question the starting index for the summation. There are attempts to express the series in terms of its components and to clarify the cancellation of terms.

Discussion Status

The discussion is ongoing, with some participants affirming the telescoping nature of the series and others pointing out potential oversights in the initial setup, such as the starting index. There is no explicit consensus yet on the correctness of the initial approach.

Contextual Notes

There is a mention of the need to clarify whether the series should start at \(n=0\) or \(n=1\), which may affect the interpretation of the sum.

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Homework Statement


Find the sum of the series: [tex]\sum^{\infty}_{0}[/tex](e[tex]^{\frac{n+1}{n}}-e^{\frac{n+2}{n+1}}[/tex])


Homework Equations


Telescoping series equation.

The Attempt at a Solution


Starting with n=0, I get: (e[tex]^{2}[/tex]-e[tex]^{\frac{3}{2}}[/tex])+(e[tex]^{\frac{3}{2}}-e^{\frac{4}{3}})+(e^{\frac{n+1}{n}}-e^{\frac{n+2}{n+1}}[/tex])=e[tex]^{2}[/tex] as n approaches infinity. Is this correct?
 
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(e^2-e^4/3)
+
(e^4/3 - e^5/4)
+
(e^5/4 - e^6/5)
+
...
+e^(n+1/n - e^(n+2)/(n+1))

=
e^2 - e^(n+2)/(n+1) <-- you forgot this.

And should it be starting with n =1?
 
Yes, that's a "telescoping" series- every term except the first cancels a later term. Well done.
 
HallsofIvy said:
Yes, that's a "telescoping" series- every term except the first cancels a later term. Well done.

Alright! Thanks!
 

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