Sum of Series with Telescoping Equation

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The discussion focuses on finding the sum of the series ∑(e^(n+1)/n - e^(n+2)/(n+1)) using the telescoping series method. The initial calculations start with n=0, leading to a series of terms that ultimately simplify to e^2 as n approaches infinity. Participants confirm that the series is indeed telescoping, where each term cancels with a subsequent term, leaving only the first term. There is a suggestion to start the series from n=1 for clarity. The conclusion affirms the correctness of the telescoping series approach in solving the problem.
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Homework Statement


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


Homework Equations


Telescoping series equation.

The Attempt at a Solution


Starting with n=0, I get: (e^{2}-e^{\frac{3}{2}})+(e^{\frac{3}{2}}-e^{\frac{4}{3}})+(e^{\frac{n+1}{n}}-e^{\frac{n+2}{n+1}})=e^{2} 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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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