indicate wether the series converges or diverges. if it converges find the sum. `sum_(n=1)^oo` `(e^n)/(e^(n-1))`



2 Answers | Add Yours

oldnick's profile pic

Posted on (Answer #1)

`sum_(n=1)^oo e^n/e^(n-1)` `=sum_(n=1)^oo e^(n-(n-1))` `=sum_(n=1)^oo e^(n-n+1)` `=sum_(n=1)^oo e` `=lim_(n->oo) n"e" = oo`

Series diverges.

justaguide's profile pic

Posted on (Answer #2)

The series is denoted by `sum_(n=1)^oo (e^n/(e^(n-1)))`

`sum_(n=1)^oo (e^n/(e^(n-1)))`

= `sum_(n=1)^oo e^(n-n+1)`

= `sum_(n=1)^oo e`

= `e + e + e + e + ...`

= `oo`

As the value of `sum_(n=1)^oo (e^n/(e^(n-1)))` is `oo` , the series is divergent.

We’ve answered 397,002 questions. We can answer yours, too.

Ask a question