1. ## Telescoping series

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

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’…

$\sum_{n=1}^{\infty}$ $(e^{\frac{4}{n}} - e^{\frac {4}{n+1}}) = e^{4} - e^{\frac{4}{2}} + e^{\frac{4}{2}} - e^{\frac{4}{3}} + e^{\frac{4}{3}} - e^{\frac{4}{4}} + \dots$

… and the value of the sum is quite evident…

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
The only you have to do is writing the series ‘term by term’…

$\sum_{n=1}^{\infty}$ $(e^{\frac{4}{n}} - e^{\frac {4}{n+1}}) = e^{4} - e^{\frac{4}{2}} + e^{\frac{4}{2}} - e^{\frac{4}{3}} + e^{\frac{4}{3}} - e^{\frac{4}{4}} + \dots$

… and the value of the sum is quite evident…

Kind regards

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

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

Kind regards

$\chi$ $\sigma$

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!

$\chi$ $\sigma$

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!

$\chi$ $\sigma$
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!

$\chi$ $\sigma$

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

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

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

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

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

8. Originally Posted by Chris L T521

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

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

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

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

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

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

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

… so that is…

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

So the series converges and its sum is what you have computed: $e^{4} -1$. 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

$\chi$ $\sigma$