Calculate the interval of convergence.
Sigma from n=1 to infinity of (n!*x^n)/n^n.
try using the ratio test. You should be able to show that the series converges for |x| < e
I tried and I got
|x|lim n-infinity of |(n/n+1)^n|
which I get |x|<1
To find the interval of convergence for the series ∑ (n!*x^n)/(n^n) from n=1 to infinity, we can use the ratio test.
The ratio test states that if the limit of the absolute value of the term-to-term ratio is less than 1, then the series converges. If the limit is greater than 1, the series diverges. And if the limit is exactly 1, the test is inconclusive.
Let's apply the ratio test to our series:
Step 1: Take the absolute value of the (n+1)th term divided by the nth term:
|((n+1)! * x^(n+1)) / ((n+1)^(n+1))| / (n! * x^n) / (n^n)
Step 2: Simplify the expression by canceling some terms:
|((n+1)! * x^(n+1)) / ((n+1)^(n+1))| * (n^n) / (n! * x^n)
Step 3: Rearrange the expression:
[((n+1)! * x^(n+1)) * (n^n)] / [(n! * x^n) * ((n+1)^(n+1))]
Step 4: Cancel out common terms:
|x * (n+1) / (n+1)^(n+1)|
Step 5: Simplify the expression:
|x / (n+1)^n|
Now, we will take the limit of this expression as n approaches infinity:
lim (n→∞) |x / (n+1)^n|
In order to evaluate this limit, we can use the limit rules:
lim (n→∞) (x / (n+1)^n) = x * (lim (n→∞) (1 / (n+1)^n))
We can further simplify this by recognizing that:
lim (n→∞) (1 / (n+1)^n) = 1 / e, where e is the mathematical constant approximately equal to 2.71828.
Therefore:
lim (n→∞) (x / (n+1)^n) = x / e
Now, since we need the limit of this expression to be less than 1 for convergence, we have:
|x / e| < 1
Taking the absolute value of x/e, we find:
|x| < |e|
So, the interval of convergence for our series is:
-|e| < x < |e|
This means that the series converges when x is between -|e| and |e|.
Note: In this case, we assumed n*! means factorial(n), and n^m means n raised to the power of m.