# Limits of series

• Apr 7th 2010, 12:06 AM
monster
Limits of series
having trouble finding limits of series,

I'm able to use ratio test/root test /integral test when told to but when presented with a series that doesn't seem to require one of these test i'm lost,

SUM(form n=1 --> n=inf) 1/((4n-3).(4n+1)),

im not sure if i need to use a test here or some other way, i tried using the ratio test but it came out to be =1 so inconclusive.

Any guidance here would be greatly appreciated,

Cheers.
• Apr 7th 2010, 12:17 AM
Drexel28
Quote:

Originally Posted by monster
having trouble finding limits of series,

I'm able to use ratio test/root test /integral test when told to but when presented with a series that doesn't seem to require one of these test i'm lost,

SUM(form n=1 --> n=inf) 1/((4n-3).(4n+1)),

im not sure if i need to use a test here or some other way, i tried using the ratio test but it came out to be =1 so inconclusive.

Any guidance here would be greatly appreciated,

Cheers.

Hint: Almost all of these kind of sums are telescoping. Use partial fractions and see what pops out.
• Apr 7th 2010, 12:20 AM
Prove It
Quote:

Originally Posted by monster
having trouble finding limits of series,

I'm able to use ratio test/root test /integral test when told to but when presented with a series that doesn't seem to require one of these test i'm lost,

SUM(form n=1 --> n=inf) 1/((4n-3).(4n+1)),

im not sure if i need to use a test here or some other way, i tried using the ratio test but it came out to be =1 so inconclusive.

Any guidance here would be greatly appreciated,

Cheers.

Try using partial fractions...

$\frac{1}{(4n-3)(4n+1)} = \frac{A}{4n - 3} + \frac{B}{4n + 1}$

$= \frac{A(4n + 1) + B(4n- 3)}{(4n - 3)(4n + 1)}$.

Therefore $A(4n + 1) + B(4n - 3) = 1$

$4An + A + 4Bn - 3B = 1$

$(4A + 4B)n + A - 3B = 0n + 1$.

Therefore $4A + 4B = 0$ and $A - 3B = 1$.

From the first equation we can see $A = -B$.

Substituting into the second equation

$-B - 3B = 1$

$-4B = 1$

$B = -\frac{1}{4}$.

So $A = \frac{1}{4}$.

So we can write

$\frac{1}{(4n - 3)(4n + 1)} = \frac{1}{4(4n - 3)} - \frac{1}{4(4n + 1)}$.

Therefore

$\sum_{n = 1}^{\infty} \frac{1}{(4n - 3)(4n + 1)} = \frac{1}{4}\sum_{n = 1}^{\infty}\left(\frac{1}{4n - 3}\right) - \frac{1}{4}\sum_{n = 1}^{\infty}\left(\frac{1}{4n + 1}\right)$

$= \frac{1}{4}\left[\sum_{n = 1}^{\infty}\left(\frac{1}{4n - 3}\right) - \sum_{n = 1}^{\infty}\left(\frac{1}{4n + 1}\right) \right]$

$= \frac{1}{4}\left[\left(1 + \frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \frac{1}{17} + \dots\right) - \left(\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + \frac{1}{17} + \dots\right)\right]$

$= \frac{1}{4}\left(1\right)$

$= \frac{1}{4}$.