# Homework Help: Find sum of convergent series: 2/[(4n-3)(4n+1)]

1. Apr 6, 2013

### Lo.Lee.Ta.

1. Find the sum of the convergent series:

Ʃ 2/[(4n-3)(4n+1)]
n=1

2. Hm... Okay, so I started with the nth term test, and the denominator gets huge very fast. So I'm pretty sure it goes to zero.

So that tells us nothing other than that it does not FOR SURE diverge.

Since it has no n in the exponent or !, I'm not doing the ratio/root test.

Since it does have a log in it, I should do the p-series comparison test.

This is where I'm having trouble.

I rewrote the problem as:

Ʃ 2/(4n2 + 8n -3) →L'Hospital's Rule→ 0/(8n + 8)
n=1
____________________________________________________________________________
I was advised that by doing L'Hospital's rule, I could figure out which p-series to compare it to.

Ʃ ln(n/n+2)
n=1

I could do L'Hospital's Rule and say that it's: [(1/n) - 1/(9n+1)]/(1/n2)

I get the 1/n2 in the denominator because there are 2 n's in the numerator.
Therefore, I have to think about what would give me the 1/n2 value when L'Hospital's rule is used.

Then I know that -1/n would give me 1/n2 when L'Hospital's rule is used, and that is also the p-series that I compare to.

Since the exponent of the n is 1, it is a divergent series.
____________________________________________________________________________

The problem is that in my problem, when using L'Hospital's Rule, I get 0/(8n + 8).

With the zero on top, I don't know what to compare it to! If it was instead a 1, I would think to compare it to 1/n...
...but the antiderivative of 1/n would be ln|n|, which is not a p-series! So I think in this case another test is required! :/

But I don't even think I can use 1/n.
Would I compare it to 0??? And what would the p-series then be???

#=_= UGH. I don't really know, as you can see! :/

Thank you so much for helping! :)

Last edited: Apr 6, 2013
2. Apr 6, 2013

### Ray Vickson

You are working on the wrong problem: you were not asked to prove convergence; you were asked to find the sum.

3. Apr 6, 2013

### jbunniii

First of all, if $n \geq 1$, then $4n-3 \geq n$ and $4n+1 \geq n$, so
$$\frac{1}{(4n-3)(4n+1)} \leq \frac{1}{n^2}$$
so the series converges by comparison with $\sum_{n=1}^{\infty}\frac{1}{n^2}$.

The various techniques you are trying are not going to help you find what number the series converges to. To do that, try writing
$$\frac{2}{(4n-3)(4n+1)}$$
in terms of partial fractions and hope that you will get telescoping terms.

4. Apr 7, 2013

### Lo.Lee.Ta.

Okay, so I need to do partial fractions here?

When I do partial fractions, A= 7/8 and B= -3/8

7/8(4n + 1) + -3/8(4n - 3)

= 2n + 2

∫2n + 2 = 2n2 + 2n

I'm confused as to what you mean by writing it in terms of partial fractions...
I guess I know how to do partial fractions, but what am I supposed to do with that...?

I don't really know how to find the sum... Help? :/

Thanks so much! :D

5. Apr 7, 2013

### jbunniii

I'm not sure what you're doing here. First of all, your $A$ and $B$ are wrong:
$$\frac{2}{(4n-3)(4n+1)} = \frac{1}{2}\left(\frac{1}{4n-3} - \frac{1}{4n+1}\right)$$
So what can you do with this? Certainly there is no point integrating it. Instead, try writing out the partial sum of the first few terms and see if you can conclude anything. In other words, if
$$s_N = \sum_{n=1}^{N} \frac{2}{(4n-3)(4n+1)}$$
then what are $s_1$, $s_2$, $s_3$? From this, can you see what will happen as $N \rightarrow \infty$?

6. Apr 7, 2013

### Lo.Lee.Ta.

*0_0 I just did partial fractions again for this problem, and I have no idea how I got that other answer!

Okay, now I get A= 1/2 and B=-1/2

Putting that back into the equation, it's now: 1/[2(4n - 3)] - 1/[2(4n + 1)]

That's the same thing as what you have when 1/2 is factored out.

1/2[(1/(4n - 3)) - (1/(4n + 1))]

So, now I should write out s1, s2, s3, ...

s1 = 1/2[(1/1) - (1/5)]

s2 = 1/2[(1/5) - (1/9)]

s3 = 1/2[(1/9) - (1/13)]

s4 = 1/2[(1/13) - (1/17)]

Hey, yay, it is telescoping!

So the 1/5, 1/9, etc. cancel out.

So I'm left with just 1/2(1) = 1/2

So... Does the sum equal 1/2?

Thanks! :D

7. Apr 7, 2013

### jbunniii

That's right. The $N$'th partial sum will be of the form
$$\frac{1}{2}\left(1 - \frac{1}{4N+1}\right)$$
and as $N \to \infty$, the fraction $1/(4N+1)$ converges to $0$, so we are left with simply $1/2$.