1. 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.

2. 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.

3. 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...

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

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

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

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

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

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

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

Substituting into the second equation

$\displaystyle -B - 3B = 1$

$\displaystyle -4B = 1$

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

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

So we can write

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

Therefore

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

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

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

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

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