# Thread: Partial Sum of a Sequence

1. ## Partial Sum of a Sequence

This is an exercise in my book that I can't figure out. And there's no answer sheet for it so I don't know if I did it right.

an = 1/n+1 - 1/n+2

I got these as the first terms of the sequence:
a1 = 1/6
a2 = 1/12
a3 = 1/20
a4 = 1/30

Based on those, I got these as the first partial sums of the sequence:
S1 = 1/6
S2 = 1/4
S3 = 3/10
S4 = 1/3

I feel the partial sums are wrong, and I can't figure out the Nth partial sum of the sequence (Sn = ?).

Any help or pointer would be greatly appreciated.

Thank you.

2. ## Re: Partial Sum of a Sequence

Originally Posted by RobinC
This is an exercise in my book that I can't figure out. And there's no answer sheet for it so I don't know if I did it right.

an = 1/n+1 - 1/n+2

I got these as the first terms of the sequence:
a1 = 1/6
a2 = 1/12
a3 = 1/20
a4 = 1/30

Based on those, I got these as the first partial sums of the sequence:
S1 = 1/6
S2 = 1/4
S3 = 3/10
S4 = 1/3

I feel the partial sums are wrong, and I can't figure out the Nth partial sum of the sequence (Sn = ?).
I am not at all sure what you are doing there.
Is it $S_N = \sum\limits_{n = 1}^N {\left( {\frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right)} ~?$
If it is then
$S_1=\frac{1}{2}-\frac{1}{3}$
$S_2=\frac{1}{2}-\frac{1}{4}$
$S_3=\frac{1}{2}-\frac{1}{5}$
$S_4=\frac{1}{2}-\frac{1}{6}$
$~ \vdots ~$
so on.

3. ## Re: Partial Sum of a Sequence

Originally Posted by Plato
I am not at all sure what you are doing there.
Yes, I'm that confused

The question states: Find the four partial sums and the nth partial sum of the sequence an.

All I'm given is: $a_n = {\left {\frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right}$

From what I read, you first find the value of $a_1, a_2, a_3, a_4$
Then you find $S_1, S_2, S_3, S_4$
So:
$S_1 = a_1$
$S_2 = a_1 + a_2$
$S_3 = a_1 + a_2 + a_3$
$S_3 = a_1 + a_2 + a_3 + a_4$

So my calculation goes like:
$S_1=\frac{1}{6}$

$S_2=\frac{1}{6} + \frac{1}{16} = \frac{1}{4}$

$S_3=\frac{1}{6} + \frac{1}{16} + \frac{1}{20} = \frac{3}{10}$

$S_4=\frac{1}{6} + \frac{1}{16} + \frac{1}{20} + \frac{1}{30} = \frac{1}{3}$

Maybe I'm doing it wrong, but this is my first math class in over 8yrs

4. ## Re: Partial Sum of a Sequence

Originally Posted by RobinC
Yes, I'm that confused
The question states: Find the four partial sums and the nth partial sum of the sequence an.
All I'm given is: $a_n = {\left {\frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right}$
From what I read, you first find the value of $a_1, a_2, a_3, a_4$
Then you find $S_1, S_2, S_3, S_4$
You are confused
Lets do $S_4$.

\begin{align*}S_4 &= a_1 + a_2 + a_3 + a_4\\ &=\left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \left( {\frac{1}{4} - \frac{1}{5}} \right) + \left( {\frac{1}{5} - \frac{1}{6}} \right)\\ &=\frac{1}{2}-\frac{1}{6}\end{align*}

Removing the parentheses all but the first and last fractions collapse.

5. ## Re: Partial Sum of a Sequence

Ah.... okay so you don't actually input the result of $a_n$ rather, you input the formula for it with the updated values for N.

So instead of: $S_2 = \left {\frac{1}{6} + \frac{1}{12}} \right$

We use: $S_2 = \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right)$

So $S_2 = \left {\frac{1}{2} - \frac{1}{4}} \right$

6. ## Re: Partial Sum of a Sequence

Originally Posted by RobinC
Ah.... okay so you don't actually input the result of $a_n$ rather, you input the formula for it with the updated values for N.

So instead of: $S_2 = \left {\frac{1}{6} + \frac{1}{12}} \right$

We use: $S_2 = \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right)$

So $S_2 = \left {\frac{1}{2} - \frac{1}{4}} \right$
That is correct. These are very famous and useful series in mathematics. They are known as collapsing sums .
Once you have the final form, then combine.

Thus, $S_n=\frac{1}{2}-\frac{1}{n+2}=\frac{n}{2n+4}$

7. ## Re: Partial Sum of a Sequence

Thank you so much for your help! It makes a lot more sense now