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

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) .