# Find the interval of absolute convergence for the given series `sum_(n=1)^(oo) e^n x^(n-1)`

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### 1 Answer

To determine the interval of absolute convergence, we need to investigate

` lim_(n->oo) (|a_(n+1)|)/(|a_n|)`

For us,

`|a_(n+1)|= e^(n+1) |x|^n`

`|a_n| = e^n |x|^(n-1)`

(we don't need absolute values around the e stuff because a positive number raised to any power is positive)

Thus:

`(|a_(n+1)|)/(|a_n|)= (e^(n+1) |x|^n)/(e^n |x|^(n-1))`

`=e|x|`

So

` lim_(n->oo) (|a_(n+1)|)/(|a_n|) = lim_(n->oo) e|x| = e|x|`

If `e|x|<1` , the series converges absolutely

If `e|x|>1` , the series diverges

If `e|x|=1`, the series might converge absolutely, might converge conditionally, or might diverge.

If `e|x|<1`, then `|x|<1/e` , so `-1/e < x < 1/e`

If `e|x|>1`, then `|x|>1/e` , so `x< -1/e` or `x> 1/e`

If `e|x|=1`, then `|x|=1/e` , so `x=1/e` or `x=-1/e`

Now we have to check explicitly the cases `x=1/e` and `x=-1/e`

Suppose `x=1/e`

Then our series becomes:

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

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

Every term in the series is e. Thus the terms don't go to 0, and we can use the test for divergence to say that this doesn't converge.

Suppose `x=-1/e`

Then our series becomes:

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

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

Again, the terms don't go to 0

Thus, neither `x=1/e` nor `x=-1/e` is in the interval of convergence, so they are not in the interval of absolute convergence.

Thus, the interval of absolute convergence is just

`-1/e < x < 1/e`