find the sum of the series: sum from k=1 to infinity [(3/k(k+1)(k+2))+(-2/3)^k]
i know that for 3/k(k+1)(k+2) I would use partial fraction representation
and the (-2/3)^k part is just a geometric series
S = (-2/3)/(1+2/3) = -2/5
To find the sum of this series, we first need to determine if it converges or diverges. If it converges, we can then find the sum.
Let's start by looking at the first term of the series: (3/k(k+1)(k+2)). One approach is to determine the limit as k approaches infinity of this term. If the limit is non-zero, then the series diverges. If the limit is zero, we need to further investigate.
Taking the limit:
lim(k→∞) (3/k(k+1)(k+2))
= lim(k→∞) (3/(k^3 + 3k^2 + 2k))
= 0
Since the limit is zero, we proceed to the next step.
Now, let's examine the second term: (-2/3)^k. This term is a geometric series. To determine if it converges or diverges, we need to check the absolute value of the common ratio.
The absolute value of the common ratio is |(-2/3)| = 2/3 < 1. Therefore, the geometric series converges.
Now that we have established that both terms converge, we can add them together to find the sum of the series.
Sum = sum of the first term + sum of the second term
The first term can be rewritten using partial fractions as follows:
3/k(k+1)(k+2) = -3/(k) + 6/(k+1) - 3/(k+2)
Now, let's express the series in terms of two separate sums:
Sum = sum from k=1 to infinity of (-3/k) + sum from k=1 to infinity of (6/(k+1)) - sum from k=1 to infinity of (3/(k+2)) + sum from k=1 to infinity of (-2/3)^k
The sum of (-3/k) is a well-known series called the harmonic series, which is known to diverge.
The sum of (6/(k+1)) can be computed by shifting the index of the sum by 1 and treating it as a telescoping series. Notice that:
sum from k=1 to infinity of (6/(k+1)) = sum from k=2 to infinity of (6/k) - sum from k=1 to infinity of (6/k)
The two terms cancel each other in the telescoping series, leaving us with:
sum from k=1 to infinity of (6/(k+1)) = 6
Similarly, the sum of (-3/(k+2)) cancels out in the telescoping series:
sum from k=1 to infinity of (3/(k+2)) = 3
Finally, the sum of the geometric series can be calculated using the formula for the sum of an infinite geometric series:
sum from k=1 to infinity of (-2/3)^k = (-1)/(1-(-2/3)) = (-1)/(5/3) = -3/5
Putting it all together:
Sum = (-∞) + 6 - 3 - 3/5
Sum = -∞ + 27/5
Therefore, the sum of the given series is -∞ + 27/5.