# Infinite Series & Limits

1. May 8, 2007

### kingwinner

1) Determine whether the infinite series

Sigma (k^2-1) / (3k^4 + 1)
k=0
converges or diverges.

[My immediate thought was to use the "limit comparsion test", but this test requires all terms to be positive. However, the first term (put k=0) is definitely negative...what should I do? Can I still use the limit comparsion test, and if not, what other tests can I use?]

2) Evaluate
lim [t^2 - t^3 sin(1/t)]
t->∞

[When I direct substitute, I get ∞-∞*0, and I have no clue how to solve this problem...any hints?]

2. May 8, 2007

### Kuno

let $$a_{n} = \frac{k^2 - 1}{3k^4 + 1}$$
and $$b_{n} = \frac{1}{k^2}$$
then show that $$a_{n} > b_{n}$$ for every n

Okay, I finally got it...

Last edited: May 8, 2007
3. May 8, 2007

### kingwinner

But if an>bn and bn converges, it tells you nothing whether an converges or not...how does that help?
And also, it doesn't look very straight foward to me to prove that an>bn

4. May 8, 2007

### kingwinner

1) Also, I am still wondering whether I can apply "limit comparsion test" to this problem or not...does anyone know?

Thanks!

5. May 8, 2007

### Office_Shredder

Staff Emeritus
This is where convergence is beautiful. It doesn't matter if the first term is negative; just look at the sum

term one + series starting from term two.

Clearly this sum exists iff your original series converges (if that's not obvious, prove it)

BTW, I think $$a_n<b_n$$ as posted above, because $$a_n$$ approximates $$\frac{1}{3k^2}$$

(in fact, it's always less than that.... you should be able to show that too)

6. May 9, 2007

### kingwinner

So when I am trying to see whether an infinite series converges or diverges, I can always change the lower index of summation whenever I want?

an approximates 1/3k^2, but how can I know which one is larger?

Thanks!

7. May 9, 2007

### quasar987

That's right; convergence or divergence of a series is independent or the starting index.