Summation Problem: Evaluate k2-k+1/k(k-1)

[darkvalentine]

Homework Statement

Evaluate: Sum[k²-k+1/k(k-1),{k,2,infinity}]

Homework Equations

The Attempt at a Solution

k²-k+1/k(k-1) can be written as k/(k-1) - 1/k, but then I get stuck because when n->infinity, the sum is divergent.

 

[Inferior89]

Your notation is slightly ambiguous.

If this is your problem:
http://www.wolframalpha.com/input/?i=Sum+k^2-k%2B1/k(k-1)+from+2+to+infinity
then yes it diverges.

 

[statdad]

If the general term is

[tex]
\frac{k^2 - k + 1}{k(k-1)}
[/tex]

the key is that numerator and denominator are polynomials of the same degree: what is the limit of this term (not the sum, just this term) as [itex] k \to \infty [/itex]?

 

[darkvalentine]

If the general term is

[tex]
\frac{k^2 - k + 1}{k(k-1)}
[/tex]

the key is that numerator and denominator are polynomials of the same degree: what is the limit of this term (not the sum, just this term) as [itex] k \to \infty [/itex]?

Thanks for input sir,
The limit of the term above when k->infinity is 1, but can I conclude that the lim(the sum above, k->infinity) = 1 also?

 

[Inferior89]

Now that I know what problem you need answer to I can give an answer. Instead of writing k2-k+1/k(k-1) write (k^2-k+1)/(k(k-1)) because the first one is not the same as the second one.

If you have
[tex] \sum_{n=0}^\infty a_n [/tex]
there is a specific property that [tex] a_n[/tex] needs to have when [tex] n \to \infty [/tex] for the sum to converge. Does your [tex] a_n [/tex] have this?

Tip: you just showed that it didn't.

 

[JonF]

Thanks for input sir,
The limit of the term above when k->infinity is 1, but can I conclude that the lim(the sum above, k->infinity) = 1 also?

Are you familiar with the limit comparison test? If so compare it to B_n = 1. This should also make it apparent that for ΣA_n to converge it must be the case that: A_n → 0 as n → ∞

 

[statdad]

As others have noted, you know that the general term does not go to zero. What does that alone say about the series' convergence?

 

[darkvalentine]

Thanks guy, I think I figure it out ))