Does Adjusting the Index Affect the Convergence of an Infinite Series?

[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?]


Thank you for your help!

 

[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...

 

[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

 

[kingwinner]

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

Thanks!

 

[Office_Shredder]

[My immediate thought was to use the "limit comparsion test", but this test requires all terms to be positive.

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)

 

[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!

 

[quasar987]

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?

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