Finding the Sum of a Convergent Series with Partial Fractions

In summary, you need to find the sum of a series if the limit is not 0. If the limit is 0, the series is divergent and cannot be computed.
  • #1
jwxie
281
0

Homework Statement



Determine whether the series converges or diverges. For convergent series, find the sum of the series.

sima (k=1, infinity), (2k +1) / ((k^2) (k+1)^2 )

Homework Equations



The Attempt at a Solution


Well, the kth test for divergence said this series has limit ak = 0 because if we simplify the leading terms, we have 2k / c*k^4, which is 1/k^3, this is a p-series. We know that for p > 1, the p-series will converge.

But how do you find the sum?
Partial fraction expanision seems not a good choice?

(2k +1) / ((k^2) (k+1)^2 ) = A/k + B/k^2 + C/(k+1) + D/(k+1)^2
->>>>
(2k+1) = A(k)(k+1)^2 + B(k+1)^2 + C(k^2)(k+1) + D(k^2)

We can ignore anything that has power higher than 1, so after mulitiplication, we have
2k + 1 = Ak + 2Bk + B

it's clear that b = 1
2k = k(a+2b)
2k = k(a+2)
2 = a + 2
a = 0

so are we left with the expansion 1/k^2 ? which doesn't make sense to me...

well anyhow, even if i have a compute error, this is not a telescopic series, no positive terms will cancel. How do I compute the sum then?
 
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  • #2
jwxie said:

Homework Statement



Determine whether the series converges or diverges. For convergent series, find the sum of the series.

sima (k=1, infinity), (2k +1) / ((k^2) (k+1)^2 )

Homework Equations



The Attempt at a Solution


Well, the kth test for divergence said this series has limit ak = 0
So the k-th term test for divergence tells us nothing.
jwxie said:
because if we simplify the leading terms, we have 2k / c*k^4, which is 1/k^3, this is a p-series. We know that for p > 1, the p-series will converge.

But how do you find the sum?
Partial fraction expanision seems not a good choice?

(2k +1) / ((k^2) (k+1)^2 ) = A/k + B/k^2 + C/(k+1) + D/(k+1)^2
->>>>
(2k+1) = A(k)(k+1)^2 + B(k+1)^2 + C(k^2)(k+1) + D(k^2)

We can ignore anything that has power higher than 1, so after mulitiplication, we have
2k + 1 = Ak + 2Bk + B

it's clear that b = 1
2k = k(a+2b)
2k = k(a+2)
2 = a + 2
a = 0

so are we left with the expansion 1/k^2 ? which doesn't make sense to me...

well anyhow, even if i have a compute error, this is not a telescopic series, no positive terms will cancel. How do I compute the sum then?
 
  • #3
wait. kth term test said if lim ak =/= 0, it must diverges. if it is zero, it may or may not diverge.

So i just that to begin with. Now I need a different approach to find the sum. It converges to 1. I mean I can approximate it goes goes to 1 if I look at the pattern numbers. But definitely I can't be sure.

How would I compute it?
 
  • #4
jwxie said:

Homework Statement



Determine whether the series converges or diverges. For convergent series, find the sum of the series.

sima (k=1, infinity), (2k +1) / ((k^2) (k+1)^2 )

Homework Equations



The Attempt at a Solution


Well, the kth test for divergence said this series has limit ak = 0 because if we simplify the leading terms, we have 2k / c*k^4, which is 1/k^3, this is a p-series.
Yes, lim ak = 0, which means that the k-th term test does not apply. You can't just simplify the leading terms, and this series is not a p-series.
jwxie said:
We know that for p > 1, the p-series will converge.

But how do you find the sum?
Partial fraction expanision seems not a good choice?

(2k +1) / ((k^2) (k+1)^2 ) = A/k + B/k^2 + C/(k+1) + D/(k+1)^2
->>>>
(2k+1) = A(k)(k+1)^2 + B(k+1)^2 + C(k^2)(k+1) + D(k^2)

We can ignore anything that has power higher than 1,
?
Carry this out. Find A, B, C, and D.
jwxie said:
so after mulitiplication, we have
2k + 1 = Ak + 2Bk + B
No, this is not how it works. On the right side you are going to have terms up to degree 3. The coefficients of the k3 and k2 terms have to be zero, but you can't just ignore them.
jwxie said:
it's clear that b = 1
2k = k(a+2b)
2k = k(a+2)
2 = a + 2
a = 0

so are we left with the expansion 1/k^2 ? which doesn't make sense to me...

well anyhow, even if i have a compute error, this is not a telescopic series, no positive terms will cancel. How do I compute the sum then?

If you complete your partial fractions work, you will find that this is a telescoping series.

jwxie said:
wait. kth term test said if lim ak =/= 0, it must diverges. if it is zero, it may or may not diverge.
And this tells us exactly nothing.
jwxie said:
So i just that to begin with. Now I need a different approach to find the sum. It converges to 1. I mean I can approximate it goes goes to 1 if I look at the pattern numbers. But definitely I can't be sure.

How would I compute it?
 

1. What is the meaning of "Find the sum of this series"?

The phrase "Find the sum of this series" refers to the process of adding up all the terms in a given series to determine the total sum.

2. How do you find the sum of a series?

To find the sum of a series, you need to add up all the terms in the series. This can be done using a formula or by hand, depending on the complexity of the series.

3. What is the formula for finding the sum of a series?

The formula for finding the sum of a series depends on the type of series. For example, for an arithmetic series, the formula is Sn = (n/2)(a + l), where n is the number of terms, a is the first term, and l is the last term. For a geometric series, the formula is S = a(1-r^n)/(1-r), where a is the first term, r is the common ratio, and n is the number of terms.

4. Can all series be summed?

No, not all series can be summed. Some series, such as divergent series, do not have a finite sum and therefore cannot be summed.

5. Why is it important to find the sum of a series?

Finding the sum of a series is important because it can help us understand the behavior and patterns in a given set of numbers. It also allows us to make predictions and solve real-world problems, such as calculating compound interest or determining the total cost of a project.

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