Find the limit of a sum

[mortal plaza]

So far I've been finding limits using the squeeze theorem, however, I am not sure what I should do when I have a sum like this. My work so far is presented in the chat.

Here is my work so far, my answer is 0. My reasoning is that since we use the squeeze theorem, we take the biggest and the smallest expression we can have from the sum and find the limit of it, since they both don't contain k, we can write it without the summation.

My questions are:

1. Is my answer and approach correct?
2. Are there any details and pitfalls I should be aware of in this kind of problems?

[wooden hinge]

you can't swap sum and limit because the sum has a number of terms that is dependent on n

[mortal plaza]

Oh yes, It's a careless mistake, didn't even see that

[wooden hinge]

Do you need a hint?

[mortal plaza]

About what? Is my answer incorrect?

[wooden hinge]

About how to solve the question

[wooden hinge]

I think you got the limit right but the justification is incorrect

[mortal plaza]

I need help with the justification then

[wooden hinge]

Then you need a different approach

[mortal plaza]

oh?

[wooden hinge]

I'll give you a hint

Let's say $S_n = \sum_{k=1}^{n} \frac{1}{2^k \sqrt{n^2+k}}$

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

Hint:
$$S_n = \frac{1}{n} \sum_{k=1}^{n} \frac{1}{2^k \sqrt{1 + \frac{k}{n^2}}}$$

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

Now this new sum is still dependent on n, both in the terms and the number of terms

but you can try to get some bounds

@mortal plaza

[mortal plaza]

Okay, but I wonder, what's wrong with the current bounds that I have? Why do I have to take this approach?

They are still less or greater than bn

[wooden hinge]

your approach is not correct because here you are trying to swap limit and sum without justification

you're arguing that the terms converge to zero so the sum of those terms must converge to zero

but the number of terms is still dependent on n

[mortal plaza]

Okay, so from there on I will be looking for the bounds, for k=1 and for k=n?

Alright, I got what you mean, for the lower bound when k=n it's alright, but for the upper bound it is not since we have a summation and I can't just take the limit of it since when k=1 we continue up to k=n because of the summation, right?

[wooden hinge]

It suffices to prove that the sum is bounded by a constant independent of n

$\exists C \geq 0, \forall n \in \bN, \sum_{k=1}^{n} \frac{1}{2^k \sqrt{1 + \frac{k}{n^2}}} \leq C$

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

Here, one useful observation is that since all the terms are positive, $\sum_{k=1}^{n} \frac{1}{2^k \sqrt{1 + \frac{k}{n^2}}} \leq \sum_{k=1}^{\infty} \frac{1}{2^k \sqrt{1 + \frac{k}{n^2}}}$

[slate jasper]

Cleanse general by deleting it

[mortal plaza]

this is what I came up with

[wooden hinge]

The justification is not really good but the idea is there

You notice that $0 \leq \sum_{k=1}^{n} \frac{1}{2^k \sqrt{n^2+k}} \leq \frac{1}{n} \sum_{k=1}^n \frac{1}{2^k} = \frac{1}{n} (1-2^{-n})$

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

However the limit of $\frac{1}{n} (1-2^{-n})$ is, as you mentioned, $0$

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

Therefore, the squeeze theorem ensures that ...
(fill in the conclusion)

[mortal plaza]

if the upper and the lower bound are equal to the same limit, the one in the middle is also equal to it?

[wooden hinge]

if a_n <= c_n <= b_n and a_n and b_n converge to the same limit, then c_n converges, and to the same limit

[mortal plaza]

I see, it is not enough to just give an answer like that, I should be more specific

[wooden hinge]

If the question requires you to prove things, then yes, you ought to be specific in what you say

[mortal plaza]

I see, I will try to do so from now on, we will be learning about proofs next semester, for this one we just got induction and that's it

I will make my answer more specific and send it here

[mortal plaza]

The language might be different but I think the idea is clear

[wooden hinge]

For the lower bounding sequence you can just pick the sequence where all the terms are equal to zero

[mortal plaza]

how so?

[wooden hinge]

Well first of all why are your c_n and b_n equal?

[mortal plaza]

Not sure, maybe if c_n has k=1 will be the right way?

[wooden hinge]

Let's try to polish the writing a bit

First: let $c_n = \sum_{k=1}^{n} \frac{1}{2^k \sqrt{n^2 + k}}$. We want to show that $c_n$ converges to $0$

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

Step 1:

Observe that $c_n \geq 0$. This should be obvious, but you can prove it. So we're gonna take $(a_n)$ the sequence such that for all $n$, $a_n = 0$.

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

Step 2:

Let $b_n = \frac{1}{n} (1 - 2^{-n})$. You must show that $c_n \leq b_n$, and that $b_n$ converges to $0$.

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

Step 3:

You now have that for all $n$, $a_n \leq c_n \leq b_n$, where $a_n$ and $b_n$ converge to 0 ($a_n$ converges to $0$ because it is always equal to $0$). What do you conclude?

[slate jasper]

Cleanse general by deleting it

[wooden hinge]

@mortal plaza

[acoustic tinsel]

@mortal plaza

:HelpIcon:| Help Reminder

Hello catxrine, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.