Find an expression for the nth partial sum using this identity

[5hassay]

Homework Statement

The series,

1 / (1 x 2) + 1 / (2 x 3) + 1 / (3 x 4) + ... + 1 / [n(n + 1)] + ...

is not a geometric series.

(A) Use the identity

1 / [k(k + 1)] = 1 / k - 1 / (k + 1)

to find an expression for the nth partial sum S_n and (B) use it to find the sum of the series. (C) Also, how many terms of the series would be required so that the partial sum differs from the total sum by less than 0.001?

Homework Equations

1 / [k(k + 1)] = 1 / k - 1 / (k + 1)

The Attempt at a Solution

(A)

I do not understand how the provided identity could be used to find S_n.
However that I think that the following is true,

S_n = 1 / (1 x 2) + 1 / (2 x 3) + ... + 1 / [n(n + 1)]

I cannot seem to make an expression such as the answer, 1 - 1 / (n +1).

(B)

I understand that the limit of S_n as n approaches infinite is equal to the sum of the series, but, alas, I do not have an expression for S_n. (The answer is 1.)

(C)

From my incomplete understanding of infinite series, I would think that the partial sum S_n would have a difference of less than 0.001 with the total sum L (the limit of the infinite series), or,

S_n - L < 0.001
S_n < 0.001 + L

Here, I would think there would be manipulation of S_n to isolate n for this inequality. (The answer is 1000.)


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Much appreciation for any help! This is my first time in maybe 3 years since doing any form of obvious sequences and series, so the smallest of corrections (incorrect terminology) is completely welcomed.

 

[Pi-Bond]

How can you use the given identity to represent 1/(1x2)? 1/(2x3)? And every term in the series?

 

[5hassay]

How can you use the given identity to represent 1/(1x2)? 1/(2x3)? And every term in the series?

Hmmmm...

S₁ = 1 / (1 x 2) = 1 / [1(1 + 1)] = 1 / 1 - 1 / (1 + 1) = 1 - 1 / 2 = 1 / 2
S₂ = 1 / (2 x 3) = 1 / [2(2 + 1)] = 1 / 2 - 1 / (2 + 1) = 1 / 2 - 1 / 3 = 1 / 6
S₃ = 1 / (3 x 4) = 1 / [3(3 + 1)] = 1 / 3 - 1 / (3 + 1) = 1 / 3 - 1 / 4 = 1 / 12

.
.
.

S_n = S₁ + S₂ + ... + 1 / n - 1 / (n + 1)

And, as I see adding S₁, S₂, and so forth, the limit as n approaches infinite of is equal to 1 (it appears). So...


S_n = 1 + 1 / n - 1 / (n + 1)


And, as I understand that the limit of 1 / n as n approaches infinite is equal to 0...

S_n = 1 + 0 - 1 / (n + 1) = 1 - 1 / (n + 1)


Is this correct? Mind you, I found the number 1 by manual calculations (I don't know if there is a 'more correct' method).

 

[Pi-Bond]

Instead of evaluating the fractions, leave them as they are. Then you can write the partial sum as:

(1-1/2)+(1/2-1/3)+(1/3-1/4)+...

Can you use this form now?

 

[5hassay]

Instead of evaluating the fractions, leave them as they are. Then you can write the partial sum as:

(1-1/2)+(1/2-1/3)+(1/3-1/4)+...

Can you use this form now?

I can now see the validity of the answer 1 - 1 / (n + 1) (2)
as the nth partial sum, in that for the first term it is immediately obvious, and that in using (2) for the nth term its difference can be verified by adding the previous partial sums.

However, I recognized this by looking at the first term and seeing the solution (2), and by applying it to an nth term and then adding the partial sums before it and seeing that they are equal. Was this what I was supposed to do?

 

[Pi-Bond]

I'm not entirely sure what you are asking; what do you mean by "previous partial sums"? The nth partial sum is the (n-1)th partial sum plus the nth term by definition. You were supposed to see the validity of the nth partial sum based on the hint I think.

 

[5hassay]

I'm not entirely sure what you are asking; what do you mean by "previous partial sums"? The nth partial sum is the (n-1)th partial sum plus the nth term by definition. You were supposed to see the validity of the nth partial sum based on the hint I think.

To be honest, I not entirely sure what I am doing, and am still pretty confused.

But, what I meant but "previous partial sums" is that in using (2) for n = 1, it results in the first value of 1 / 2; for the second term n = 2, the equation gets the same value as in adding the first and second terms; and so on.

 

[vela]

You're confusing the two sequences. You have one sequencea_k = \frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}and the sequence of partial sumsS_n = \sum_{k=1}^n a_k=a_1 + a_2 + \cdots + a_nCalculate the n-th partial sum S_n by substituting in for a₁, a₂, …, a_n and using the hint. (I think this is what you may have done, but it's not clear because you're using S_n to refer to both sequences.)

 

[Pi-Bond]

To be honest, I not entirely sure what I am doing, and am still pretty confused.

But, what I meant but "previous partial sums" is that in using (2) for n = 1, it results in the first value of 1 / 2; for the second term n = 2, the equation gets the same value as in adding the first and second terms; and so on.

Yes, that is what the partial sum represents; the sum of the first n terms of a sequence. Of course, that is what you get when you evaluated-

(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-1/(n-1)) = 1- 1/(n-1)

 

[5hassay]

Well, I appreciate all the responses, but I have become a bit frustrated and am going to give this problem some time and come back to it later, and maybe I will understand these hints and help I have been given a bit better, haha.