Finding the limit

Hi,
I need to find the following limit and I am not sure if what I did is right. In the question it gives a hint to use the formula for the sum of an arithmetic progression but I did not even use it and so I am think I did something wrong. Here is what I did:


Thank you in advance for your help!

Perhaps you've attached the wrong picture(?) - never quite heard of the limit of a vector before...

Hi,
I need to find the following limit and I am not sure if what I did is right.

The limit is not zero.

You've used, in effect, $\lim a_nb_n =\lim a_n \times \lim b_n$

Having separated out the 1/n^2, the remainder is an AP (n + (n+1) +...+3n) and the limit of that AP as n tends to infinity is infinity.

And 0 x infinity is undetermined.

You need to use the formula on the AP.

Edit: Don't like how I've phrased this, but it's late....

I see what I mean. However, I cannot really see the arithmetic progression in (n+n+1+...+3n). I thought of separating the variables in the following way (1+2+3+4+...+n + n*n +...+3n) but I am not sure what is in between n+1 and 3n. Is it just terms to which we add 1, like (n+2)+(n+3)+(n+4)+... and so on?

Note that the sum doesn't start at "1", except in the very first term.

In the n'th term in the sequence, the numerators in the sum part, go up in steps of 1, thus:

$n+(n+1)+(n+2)+...+(3n-2) + (3n-1)+(3n)$

Writing out the first few terms of the sequence may help clarify things for you.

Thank you for your reply. If I understood the progression of the terms correctly (i.e. they increase by 1 until 3n), I got the following:

sum = {(2n + (n-1)*1)/2} *2n = (6n^2 - 2n)/2 = 3n^2 - n

and if I take the limit of this, then I will obtain infinity again and so the limit will be undefined

I hope I understood the AP correctly and thanks in advance!

You've run into the old "fencepost" problem, if things increase by one then the number of them is "last - first +1". I think it's easier to use the (n/2) (a+l) formula for the sum of an AP here, but you still need to be aware of the "fencepost" issue. (Although it actually makes no difference to the limit.)

Also the "n-1" in the inner brackets is incorrect and that does effect the outcome.

Correcting that, you also want to include the 1/n^2. Don't separate it out and take its limit separately. You want the whole thing together and take just one limit, once you've simplified the expression using the AP formula.

Sorry I am being really slow but I got that the AP is:
{(n+3n)*(2n+1)}/2 and so by taking the limit of {1/n^2}*{{(n+3n)*(2n+1)}/2} I get 4. Is it correct by any chance?

Yep.


Note:
Most of your problems stemed from splitting the terms in the sequence into two and taking two separate limits, one of which was zero, and the other infinite. So, ....