Summation of Infinite Series with Alternating Denominators

[jdawg]

Homework Statement

Find the sum of the series:
^∞_n=1∑(6)/((2n-1)(2n+1))

Homework Equations

The Attempt at a Solution

S₁=2
S₂=2+(6/15)
S₃=2+(6/15)+(6/35)

This is the part where I get a little confused. It looks like the denominator is getting bigger... So does it approach infinity and diverge?

 

[Saitama]

Homework Statement

Find the sum of the series:
^∞_n=1∑(6)/((2n-1)(2n+1))

Homework Equations

The Attempt at a Solution

S₁=2
S₂=2+(6/15)
S₃=2+(6/15)+(6/35)

This is the part where I get a little confused. It looks like the denominator is getting bigger... So does it approach infinity and diverge?

Hi jdawg!

I suggest decomposing into partial fractions.

 

[jdawg]

Hi jdawg!

I suggest decomposing into partial fractions.

Ohh! Thanks! I'll give that a shot :)

 

[jdawg]

(6)/((2n-1)(2n-1))=(A/(2n-1))+(B/(2n+1)

6=A(2n+1)+B(2n-1)

6=A2n+A+B2n-B

6=A-B A2n+B2n=0
A=6+B

(6+B)2n+B2n=0
12n+B2n+B2n=0
12n-2B2n

I don't think I'm doing this right

 

[Saitama]

(6)/((2n-1)(2n-1))=(A/(2n-1))+(B/(2n+1)

6=A(2n+1)+B(2n-1)

6=A2n+A+B2n-B

6=A-B A2n+B2n=0
A=6+B

(6+B)2n+B2n=0
12n+B2n+B2n=0

Correct so far.

12n-2B2n

I can't understand this step. What have you done here?

 

[jdawg]

Correct so far.

I can't understand this step. What have you done here?

Oops! Sorry, ignore that step. Could you do this?:

12n+B2n+B2n=0
B(2n+2n)=-12n
B(4n)=-12n
B=-3

A=6+B
A=6-3
A=3

 

[Saitama]

Oops! Sorry, ignore that step. Could you do this?:

12n+B2n+B2n=0
B(2n+2n)=-12n
B(4n)=-12n
B=-3

A=6+B
A=6-3
A=3

Looks right to me. :)

Can you proceed with the problem?

 

[jdawg]

Looks right to me. :)

Can you proceed with the problem?

Awesome! Ok so:
^∞_n=1 ∑((3)/(2n-1))-((3)/(2n+1))

S₁=2
S₂=2+(2/5)
S₃=2+(2/5)+(6/35)

This is the part that really confuses me, how do you find the sum?

Also, this might be a dumb question, but if you find what the series converges to is that number the sum, or is that just what the limit approaches? I don't have a very good grasp on series yet.

 

[Saitama]

Awesome! Ok so:
^∞_n=1 ∑((3)/(2n-1))-((3)/(2n+1))

S₁=2
S₂=2+(2/5)
S₃=2+(2/5)+(6/35)

Don't calculate the sums! Write the sums in the following way:
$$S_1=\frac{3}{1}-\frac{3}{3}$$
$$S_2=\frac{3}{1}-\frac{3}{3}+\frac{3}{3}-\frac{3}{5}$$
$$S_3=\frac{3}{1}-\frac{3}{3}+\frac{3}{3}-\frac{3}{5}+\frac{3}{5}-\frac{3}{7}$$
Do you see now? :)

 

[jdawg]

Don't calculate the sums! Write the sums in the following way:
$$S_1=\frac{3}{1}-\frac{3}{3}$$
$$S_2=\frac{3}{1}-\frac{3}{3}+\frac{3}{3}-\frac{3}{5}$$
$$S_3=\frac{3}{1}-\frac{3}{3}+\frac{3}{3}-\frac{3}{5}+\frac{3}{5}-\frac{3}{7}$$
Do you see now? :)

Ohhh! I see the pattern in the denominator now, it looks like its increasing by +2 each time?
Is this the part where you do S_n?
S_n=(3)/(n)-(3)/(n+2)
I don't think that's quite right though...

 

[Saitama]

Ohhh! I see the pattern in the denominator now, it looks like its increasing by +2 each time?
Is this the part where you do S_n?
S_n=(3)/(n)-(3)/(n+2)
I don't think that's quite right though...

Not correct. If the individual sums doesn't help, you can write the given summation as follows:
$$\sum_{n=1}^{\infty} \frac{3}{2n-1}-\frac{3}{2n+1}=3\left(\sum_{n=1}^{\infty} \frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
$$=3\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7} \cdots \right)$$

I hope this helps.

 

[jdawg]

Not correct. If the individual sums doesn't help, you can write the given summation as follows:
$$\sum_{n=1}^{\infty} \frac{3}{2n-1}-\frac{3}{2n+1}=3\left(\sum_{n=1}^{\infty} \frac{1}{2n-1}-\frac{1}{2n+1}\right)$$
$$=3\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7} \cdots \right)$$

I hope this helps.

Thanks so much, you were super helpful! :)