`sum_(n=0)^oo (x+1)^n/(n!)` Find the values of x for which the series converges.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`sum _(n=0)^oo (x+1)^n/(n!)`

To determine the interval of convergence, use Ratio Test.  The formula in Ratio Test is:

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

If L<1, the series is absolutely convergent. 

If L>1, the series is divergent.

And if L=1, the test is inconclusive. The series may converge or diverge.

Applying the...

See
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Get 48 Hours Free Access

`sum _(n=0)^oo (x+1)^n/(n!)`

To determine the interval of convergence, use Ratio Test.  The formula in Ratio Test is:

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

If L<1, the series is absolutely convergent. 

If L>1, the series is divergent.

And if L=1, the test is inconclusive. The series may converge or diverge.

Applying the formula above, the value of L will be:

`L= lim_(n->oo) |((x+1)^(n+1)/((n+1)!))/((x+1)^n/(n!))|`

`L=lim_(n->oo) | (x+1)^(n+1)/((n+1)!)*(n!)/(x+1)^n|` 

`L= lim_(n->oo) | (x+1)^(n+1)/((n+1)n!) *(n!)/(x+1)^n|`

`L = lim_(n->oo) | (x+1)/(n+1)|`

`L= (x+1) lim_(n->oo) |1/(n+1)|`

`L=(x+1) * 0`

`L=0`

Since the value of L is less than 1, the given series converges for all values of x.

Therefore, the interval of convergence  is `(-oo, oo)` .

Approved by eNotes Editorial Team