$\displaystyle \sum^{\infty}_{n=1} (e^{\frac{4}{n}} - e^{\frac{4}{n+1}})$
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!
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
$\displaystyle \chi$ $\displaystyle \sigma$
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, $\displaystyle \sum_{n=1}^{\infty}e^{\frac{4}{n}}-e^{\frac{4}{n+1}}=\boxed{e^4-1}$.
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
$\displaystyle \chi$ $\displaystyle \sigma$