i have to determine whether the series is convergent, and if, find the sum
the series is from k=1 to infinity of
2/((k+1)(k+3))
I got 5/6 as my answer and didn't know if it was right...
There are many tests for convergence. In this case you can compare the series to the known convergent series 1/k^2 from k = 1 to infinity. Since
2/((k+1)(k+3)) is always less than 2/k^2 the series must converge.
The summation can be computed by writing:
2/((k+1)(k+3)) =
1/(k+1) - 1/(k+3)
So, you would expect the summation to be given by the sum of 1/(k+1) from k = 1 to k = 2 (the two terms that are not conceled by 1/(k+3)), which is the answer you obtained.
Since the two separate summations do not converge you must prove that this is indeed the case. What you do is you say that the summation from 1 to infinity of 2/((k+1)(k+3)) equals the limit of N goes to infinity of the summation from 1 to N. For finite N you then evaluate the summation of
1/(k+1) - 1/(k+3)
from k = 1 to N.
You then get four terms: the k = 1 and k = 2 terms of 1/(k+1) which yields 5/6 and the k = N-1 and k = N terms of
- 1/(k+3). So, you get:
5/6 - 1/(N+2) - 1/(N+3)
Finally, you take the limit
N --> infinity which yields the answer of 5/6
The terms can be rewritten
1/(k+1) -1/(k+3)
I used the technique of partial sums. You can verify that the two expressions are equivalent.
Summing from x=1 to infinity, all terms will cancel in pairs except 1/2 + 1/3, for k=1 and k=2. The sum of those terms is 5/6, so I agree.
Did you use a different method?
kinda helpful
To determine whether the series is convergent and find its sum, we can use the method of partial fractions and the telescoping series.
First, let's rewrite the series using partial fractions. We want to express the term 2/((k+1)(k+3)) as a sum of simpler fractions. We can do this by decomposing the denominator:
2/((k+1)(k+3)) = A/(k+1) + B/(k+3).
To determine the values of A and B, we need to find a common denominator:
2 = A(k+3) + B(k+1).
Expanding and comparing coefficients, we get:
2 = (A + B)k + 3A + B.
Equating the coefficients of k on both sides gives:
A + B = 0.
Equating the constant terms on both sides gives:
3A + B = 2.
Solving this system of equations, we find A = -2/3 and B = 2/3.
Therefore, we can rewrite the original series as:
2/((k+1)(k+3)) = -2/3 * 1/(k+1) + 2/3 * 1/(k+3).
Now, let's rewrite the sum of the series using the partial fractions:
∑ (from k=1 to infinity) 2/((k+1)(k+3)) = ∑ (from k=1 to infinity) (-2/3) * 1/(k+1) + ∑ (from k=1 to infinity) 2/3 * 1/(k+3).
We can rewrite the first sum as:
∑ (from k=1 to infinity) (-2/3) * 1/(k+1) = (-2/3) * (∑ (from k=2 to infinity) 1/k).
Similarly, we can rewrite the second sum as:
∑ (from k=1 to infinity) 2/3 * 1/(k+3) = (2/3) * (∑ (from k=4 to infinity) 1/k).
Now, observe that as we expand both sums, many terms will cancel out due to the telescoping property of the series.
The telescoping property states that when we add up the terms of a telescoping series, most of the terms will cancel, leaving only a finite number of terms.
Let's expand the first sum:
(-2/3) * (∑ (from k=2 to infinity) 1/k) = (-2/3) * ((1/2) + (1/3) + (1/4) + (1/5) + ...).
We notice that many terms cancel out:
((-2/3) * (1/2)) + ((-2/3) * (1/3 - 1/2)) + ((-2/3) * (1/4 - 1/3)) + ((-2/3) * (1/5 - 1/4)) + ...
= (-2/3) * (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ...).
Similarly, let's expand the second sum:
(2/3) * (∑ (from k=4 to infinity) 1/k) = (2/3) * ((1/4) + (1/5) + (1/6) + (1/7) + ...).
Again, many terms cancel out:
((2/3) * (1/4)) + ((2/3) * (1/5 - 1/4)) + ((2/3) * (1/6 - 1/5)) + ((2/3) * (1/7 - 1/6)) + ...
= (2/3) * (1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + ...).
Now, if we observe the pattern in both expanded sums, we can see that most terms cancel out, leaving only the first term from each sum:
(-2/3) * (1/2) + (2/3) * (1/4).
Simplifying this expression, we get:
(-2/6) + (2/12) = -1/3 + 1/6 = -1/6.
Hence, the sum of the given series is -1/6.