find the sum of the series from n=1 to infinity of
2^n/n!
do I just look at the sum of the terms?
Sum from n = 0 to infinity of x^n/n! = exp(x)
So, Sum from n = 1 to infinity of x^n/n! = exp(x) - 1
To find the sum of the series from n=1 to infinity of 2^n/n!, you cannot simply look at the sum of the terms of the series. This is because the series is infinite, and the sum of an infinite series may not be immediately obvious.
Instead, you can approach this problem by using the concept of convergence of series. A series converges if its terms approach a finite limit as the number of terms goes to infinity. In this case, we can use the ratio test to determine whether the series converges or diverges.
The ratio test states that if the absolute value of the ratio of the (n+1)-th term to the n-th term approaches a value less than 1 as n goes to infinity, then the series converges. Conversely, if the ratio approaches a value greater than 1 or infinity, the series diverges.
Let's apply the ratio test to the given series:
First, let's consider the ratio of the (n+1)-th term to the n-th term:
R = (2^(n+1))/(n+1)! / (2^n)/n!
= (2^(n+1) * n!)/(n+1)! * (n!)/(2^n)
= 2(n+1)/((n+1)(n!)/(2^n))
= 2(n+1)/(n+1) * (2^n/n!)
= 2 * (2^n/n!)
The (n+1)-th term cancels out with the (n+1) term in the denominator, leaving us with:
R = 2
Since the ratio R = 2, which is greater than 1, the series diverges.
Therefore, the sum of the series from n=1 to infinity of 2^n/n! does not exist because the series diverges.