1. ## Telescoping series

$\displaystyle \sum^{\infty}_{n=1} (e^{\frac{4}{n}} - e^{\frac{4}{n+1}})$

I'm not sure how to set this up so I can evaluate it as a telescoping sum. Do I need to pull out a negative and make it a ratio? Thanks!

2. The only you have to do is writing the series ‘term by term’…

$\displaystyle \sum_{n=1}^{\infty}$$\displaystyle (e^{\frac{4}{n}} - e^{\frac {4}{n+1}}) = e^{4} - e^{\frac{4}{2}} + e^{\frac{4}{2}} - e^{\frac{4}{3}} + e^{\frac{4}{3}} - e^{\frac{4}{4}} + \dots … and the value of the sum is quite evident… Kind regards \displaystyle \chi \displaystyle \sigma 3. Originally Posted by chisigma The only you have to do is writing the series ‘term by term’… \displaystyle \sum_{n=1}^{\infty}$$\displaystyle (e^{\frac{4}{n}} - e^{\frac {4}{n+1}}) = e^{4} - e^{\frac{4}{2}} + e^{\frac{4}{2}} - e^{\frac{4}{3}} + e^{\frac{4}{3}} - e^{\frac{4}{4}} + \dots$

… and the value of the sum is quite evident…

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
I get $\displaystyle e^4$ is that correct?

4. Originally Posted by mollymcf2009
I get $\displaystyle e^4$ is that correct?
Yes!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Dear Molly…
… I’m afraid have made a great mistake!… The series you have proposed has termes with alternating sign and decreasing in absolute value. But necessary condition for convergence is that the general term tend to 0 and in this case it isn’t. So the series doesn’t converge… and your thanks to me are totally not deserved

I apologize and promit never to repeat such a terrible mistake!…

Very sorry!

$\displaystyle \chi$ $\displaystyle \sigma$

6. Originally Posted by chisigma
Dear Molly…
… I’m afraid have made a great mistake!… The series you have proposed has termes with alternating sign and decreasing in absolute value. But necessary condition for convergence is that the general term tend to 0 and in this case it isn’t. So the series doesn’t converge… and your thanks to me are totally not deserved

I apologize and promit never to repeat such a terrible mistake!…

Very sorry!

$\displaystyle \chi$ $\displaystyle \sigma$
No worries! I figured it out later after trying it that same way a few times. MEant to post corrections and forgot. Don't sweat it! Thanks for commenting on it though!

7. Originally Posted by chisigma
Dear Molly…
… I’m afraid have made a great mistake!… The series you have proposed has termes with alternating sign and decreasing in absolute value. But necessary condition for convergence is that the general term tend to 0 and in this case it isn’t. So the series doesn’t converge… and your thanks to me are totally not deserved

I apologize and promit never to repeat such a terrible mistake!…

Very sorry!

$\displaystyle \chi$ $\displaystyle \sigma$

To evaluate this telescoping series, first find the sum of the first $\displaystyle N$ terms.

Thus, $\displaystyle \sum_{n=1}^{N}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=e^{4}-e^{2}+e^2-e^{\frac{4}{3}}+\dots+e^{\frac{4}{N}}-e^{\frac{4}{N+1}}=e^4-e^{\frac{4}{N+1}}$

Now to evaluate $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}$, take the sum of the first $\displaystyle N$ terms and see what happens as $\displaystyle N\to\infty$.

In otherwords, evaluate $\displaystyle \lim_{N\to\infty}e^{4}-e^{\frac{4}{N+1}}$
Thus, $\displaystyle \lim_{N\to\infty}e^{4}-e^{\frac{4}{N+1}}=e^4-e^0=e^4-1$.

Therefore, $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=\boxed{e^4-1}$.

8. Originally Posted by Chris L T521

To evaluate this telescoping series, first find the sum of the first $\displaystyle N$ terms.

Thus, $\displaystyle \sum_{n=1}^{N}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=e^{4}-e^{2}+e^2-e^{\frac{4}{3}}+\dots+e^{\frac{4}{N}}-e^{\frac{4}{N+1}}=e^4-e^{\frac{4}{N+1}}$

Now to evaluate $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}$, take the sum of the first $\displaystyle N$ terms and see what happens as $\displaystyle N\to\infty$.

In otherwords, evaluate $\displaystyle \lim_{N\to\infty}e^{4}-e^{\frac{4}{n+1}}$.

Thus, $\displaystyle \lim_{N\to\infty}e^{4}-e^{\frac{4}{N+1}}=e^4-e^0=e^4-1$.

Therefore, $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=\boxed{e^4-1}$.
Effectively the general term of the series is…

$\displaystyle a_{n} = e^{\frac{4}{n}}-e^{\frac{4}{n-1}}$

… so that is…

$\displaystyle \lim_{n\to\infty}a_{n}=\lim_{n\to\infty} e^{\frac{4}{n}}\cdot (1-e^{\frac{4}{n+1}}\cdot e^{- \frac{4}{n}})=\lim_{n\to\infty} e^{\frac{4}{n}}\cdot (1-e^{- \frac{4}{n(n+1)}})= 0$

So the series converges and its sum is what you have computed: $\displaystyle e^{4} -1$. Tonight i discovered to have [probably] done a bad suggestion to Molly and for that i was 'furious' ... This story contains an useful teaching: hurry gives bad suggestions...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$