Is Using the Comparison Test to Prove Divergence Valid for This Series?

[arhzz]

Summary:: I am suspossed to find the limit of this series.I've come to realize that the series diverges and I'm trying to prove that using the a comparison test.

Hello!

Consider this sum

$$ \sum_{k=1}^{n} (\sqrt{1+k} - \sqrt{k}) $$ the question wants me to find the limit of this sum where n is approching infinity.My first hunch was that this series is diverging and I wanted to prove that using the comparison test.I first tried to find a diffrent series that diverges and that it is smaller than my original series so (an>bn) and prove it that way.So what I did is I expanded the denominTOR (I am not sure if that is the right term in English but I think you will get what I meant)$$
\sqrt{k+1}-\sqrt{k}=\frac{\left(\sqrt{k+1}-\sqrt{k}\right)\cdot\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}
$$
Now I have a series that is smaller than my original one,so now I need to prove that this series diverges.Since this isn't a "standard series" (harmonic,telescopic...) I cannot directly see if it diverges.So I tried "modifying" this series to one of the series in which I can easily determine if it diverges or not and go on from there.

What I did here is I've transformed the roots of the denominator like this.

$$\frac{1}{(k+1)^{1/2} + k^{1/2}} $$ Now this should be a harmonic series $\frac{1}{k^s}$ where S is the exponent.Now if the exponent is greater than 1 the series converges,if it is smaller it diverges.Hence this series is divergent,which means the original series is also divergent.According to an online calculator this is correct,but I am not sure if the methods I used are "allowed" (if the math adds up).Would anyone be able to confirm if this is legit or not? Also if not what do I need to redo?PS: I've tried looking in the "Homework help" department to post my question (since it is homework related) but I only found forums regarding physics and engineering.If this question doesn't belong here,please let me know in which forum I should post it.Thank you and excuse the long post!

[Moderator's note: moved from a technical forum.]

 

[fresh_42]

Consider $\dfrac{1}{2\sqrt{k+1}}<\dfrac{1}{\sqrt{k+1}+\sqrt{k}}<\dfrac{1}{2\sqrt{k}}.$

 

[arhzz]

Consider $\dfrac{1}{2\sqrt{k+1}}<\dfrac{1}{\sqrt{k+1}+\sqrt{k}}<\dfrac{1}{2\sqrt{k}}.$

I am not quite sure what I'm susposed to see here? I am guessing its something in the lines of the Sandwich-Lemma but I am not seeing how this proves divergence?

 

[fresh_42]

If the left and the right side diverges, how could the series in the middle converge? You only must show that $\displaystyle{\sum \dfrac{1}{\sqrt{n}}$ diverges.

 

[arhzz]

If the left and the right side diverges, how could the series in the middle converge? You only must show that $\displaystyle{\sum \dfrac{1}{\sqrt{n}}$ diverges.

Okay so basically Sandwich-Lemma. The LaTeX kinda slipped and I'm not able to tell what there really is written.But this still isn't really what I am looking for.I know taht the series is not convering,I am not trying to prove that either I am trying to prove that it diverges.I've done that but I just don't know if the method I am doing is allowed,correct whatever you want to put it.Same would be applied to whatever sum you have written out for me,how am i susposed to show it.Can I transform it so that I get a harmonic series,telescopic etc.. Or do I have to use a lim? But that just wouldn't make sense because the limit is n to infinity and we have k's in the sum.

 

[fresh_42]

If you can show that ∑ 1/√n diverges, then the Sandwich lemma does the rest.

I'm not sure which tools are allowed to use. E.g. ∑ 1/√n > ∑ 1/n which diverges by the minority criterion.

 

[arhzz]

If you can show that ∑ 1/√n diverges, then the Sandwich lemma does the rest.

I'm not sure which tools are allowed to use. E.g. ∑ 1/√n > ∑ 1/n which diverges by the minority criterion.

Well to prove that ##\frac{1}{sqrt{n}} diverges when n goes to infinity should be simple.

We can rewrite the root as 1/2 and than the series would definately be diverging since that is a harmonic series where if s < 1 the series is always diverging.I kind of did the same trick with the upper one,so I'd say that my method should be okay?

 

[fresh_42]

Sure. You didn't even rearrange the sum which is problematic in case it is not absolutely convergent. You simply wrote it in a different way. The inequalities I wrote are almost obvious so that your series is trapped between two divergent series, which makes it divergent itself.

 

[Office_Shredder]

Can I transform it so that I get a harmonic series,telescopic etc.

You do realize the original series you were given is a telescoping sum, right?

 

[arhzz]

You do realize the original series you were given is a telescoping sum, right?

To be 100% honest I kind of don't see that,and would that mean I could have "applied" the knowledge of the telescopic series here.But how,I am not too familiar with the telescopic series,how does one check for the converge or divergence in this case,and how do I work around the roots (if at all necessary).

 

[Office_Shredder]

Just write down the first five terms of the series and I think it will be obvious

 

[fresh_42]

You do realize the original series you were given is a telescoping sum, right?

<--- me

 

[FactChecker]

To be 100% honest I kind of don't see that,and would that mean I could have "applied" the knowledge of the telescopic series here.But how,I am not too familiar with the telescopic series,how does one check for the converge or divergence in this case,and how do I work around the roots (if at all necessary).

Notice that a lot of the series cancels out and the sum has a simple formula.

 

[arhzz]

Okay so $$ (\sqrt{1+1} - \sqrt{1}) - (\sqrt{2+1} - \sqrt{2}) - (\sqrt{3+1} - \sqrt{3}) $$

$$ \sqrt{2} - \sqrt{1} - \sqrt{3} - \sqrt{2} - \sqrt{4} + \sqrt{3}$$ We can see that a lot of these cancel out I stopped at 4 but I'd assume everything would ancel out except the $-\sqrt{1}$ Now how am I susposed to get if the series diverges or converges here?

 

[Office_Shredder]

You are subtracting every term instead of adding them together. And along with the $\sqrt{1}$ if you add the first n terms together there is one other term that always survives. What is it in your example? Try going out one more term and see what it is there.

 

[fresh_42]

You have to pay more attention to the signs. You can also use the abstract sums:
$$
\sum_{k=1}^n(\sqrt{k+1}-\sqrt{k})= \sum_{k=1}^n \sqrt{k+1}- \sum_{k=1}^n\sqrt{k}=\sum_{j=2}^{n+1} \sqrt{j}- \sum_{k=1}^n\sqrt{k}=\sum_{k=2}^{n+1} \sqrt{k}- \sum_{k=1}^n\sqrt{k}
$$
and now observe which terms remain. These are finite sums, so you can calculate with it like you do with any additions.

 

[arhzz]

Do I have to add them in the way between the parantheses should be a + not a minus,because that would make sence? But than how do I get $\sqrt{1}$ ?? Also I still don't see what term remains except the root 1.I've written it out till 6 and they still cancel out (if we have + only between the parantheses)

 

[fresh_42]

Do I have to add them in the way between the parantheses should be a + not a minus,because that would make sence? But than how do I get $\sqrt{1}$ ?? Also I still don't see what term remains except the root 1.I've written it out till 6 and they still cancel out (if we have + only between the parantheses)

Did you consider my formal sums? Check it for $n=2$ and $n=3$. Yes, $-\sqrt{1}=-1$ remains, but there is another term that does not cancel! Note that we have two finite sums! The convergence considerations come in a second step after you calculated the sum.

 

[Office_Shredder]

Do I have to add them in the way between the parantheses should be a + not a minus,because that would make sence? But than how do I get $\sqrt{1}$ ?? Also I still don't see what term remains except the root 1.I've written it out till 6 and they still cancel out (if we have + only between the parantheses)

If you just take one term, you get $\sqrt{2}-\sqrt{1}$. What if you take two terms? What if you take three terms? What are the actual numbers you end up with in each case?

 

[arhzz]

Did you consider my formal sums? Check it for $n=2$ and $n=3$. Yes, $-\sqrt{1}=-1$ remains, but there is another term that does not cancel! Note that we have two finite sums! The convergence considerations come in a second step after you calculated the sum.

Im sorry I cannot understand what you meant by those sums (post #16).Best I can do is the good ol'way of simply plugging it into the series I was giving at the start.

 

[Office_Shredder]

Im sorry I cannot understand what you meant by those sums (post #16).Best I can do is the good ol'way of simply plugging it into the series I was giving at the start.

Really, just write down the first couple finite sums that you get. If you add only one term from the series you get $\sqrt{2}-\sqrt{1}$. What do you get if you add the first two terms together? What do you get if you add the first three together?

 

[mfb]

Then do that. Write down the first three or four elements and add them. Note what you get as sum of the first two, then first three, and calculate the sum of the first four if you like. The pattern should be obvious.

 

[arhzz]

If you just take one term, you get $\sqrt{2}-\sqrt{1}$. What if you take two terms? What if you take three terms? What are the actual numbers you end up with in each case?

I think I finally got it
If we take two terms we get $\sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2}$
If we take three terms we get $\sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3}$

Assuming this is correct (which I am pretty sure it is)we can see that $-\sqrt{1}$ will be there and the $\sqrt{k+1}$ will remain.

 

[fresh_42]

I think I finally got it
If we take two terms we get $\sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2}$
If we take three terms we get $\sqrt{2} - \sqrt{1} + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3}$

Assuming this is correct (which I am pretty sure it is)we can see that $-\sqrt{1}$ will be there and the $\sqrt{k+1}$ will remain.

$\sqrt{n+1}-1$ remains. It is the value of the partial sum. Now you can consider $n\to \infty .$

 

[arhzz]

$\sqrt{n+1}-1$ remains. It is the value of the partial sum. Now you can consider $n\to \infty .$

Wow that took so long to figure out... Should be infinity, the limit of $\sqrt{n+1}$ is infinity,and the limit of -1 is -1,hence the limit is infinity.This means that the series is not approaching any particular value,but grows without bounds so it diverges.

 

[Office_Shredder]

Looks good to me now!

 

[arhzz]

Excellent,thank you for your help,All of you!

 

[epenguin]

? Can we just write
$$\left( 1+k\right) ^{1/2}-k^{1/2}=\left[ k\left( 1+\dfrac {1}{k}\right) \right] ^{1/2}-k^{1/2}$$
$$=k^{1/2}\left[ \left( 1+\dfrac {1}{k}\right) ^{1/2}-1\right] $$
For large $k$ the quantity in round brackets is approximately equal to, and more importantly, larger than $\left( 1+\dfrac {1}{2k}\right)$ . Thus the whole for large k is approximately equal to but greater than $\dfrac {1}{2k^{1/2}}$ . The expression is reduced to a single term.
You perhaps recognise the infinite series of such terms as divergent, or else its divergence can be proved by the integral test or otherwise.

 

[arhzz]

? Can we just write
$$\left( 1+k\right) ^{1/2}-k^{1/2}=\left[ k\left( 1+\dfrac {1}{k}\right) \right] ^{1/2}-k^{1/2}$$
$$=k^{1/2}\left[ \left( 1+\dfrac {1}{k}\right) ^{1/2}-1\right] $$
For large $k$ the quantity in round brackets is approximately equal to, and more importantly, larger than $\left( 1+\dfrac {1}{2k}\right)$ . Thus the whole for large k is approximately equal to but greater than $\dfrac {1}{2k^{1/2}}$ . The expression is reduced to a single term.
You perhaps recognise the infinite series of such terms as divergent, or else its divergence can be proved by the integral test or otherwise.

That is definately a legit way to do it,although it goes into that teritory I am not comftorable with yet,where you have to make a lot of good educated guesses.Still thank you for the insight!

 

[epenguin]

You will probably later come across

That is definately a legit way to do it,although it goes into that teritory I am not comftorable with yet,where you have to make a lot of good educated guesses.Still thank you for the insight!

You will probably some few times come across the approximation $√(1 + a) ≈ (1 + a/2)$ for small $a$ in treatments in various physics and chemistry. P But further! Always connect!

It is what you are using in the standard school arithmetical calculation of square roots!

And then again do you notice the the similarity between $\left( 1+k\right) ^{1/2}-k^{1/2}$ and the basic formula for all differentiation $[ f(x + δ) - f(x)]/δ$ ? Indeed the first formula is the second for $f(k)= √k$ and $δ = 1$. Now we usually posit that $δ$ be small - that is only in comparison with $f(k)$ so if we let $k$ be large enough, 1 is a small by comparison and our formula is a good approximation to the derivative; as we let $k$ increase to infinity it becomes the derivative.
$$\sqrt {1+k}-\sqrt {k}=\dfrac {d\sqrt {k}}{dk}$$
What about the sum of the series? We can say approximately that for some large $k$ called $K$
$$\sum ^{\infty }_{k=K}(\sqrt {1+k}-\sqrt {k})= \sum ^{\infty }_{k=K}\dfrac {d\sqrt {k}}{dk}$$
and we can approximate the sum by an integral, which is of the easiest
$$\int ^{\infty }_{K}\dfrac {d\sqrt {k}}{dk}.dk=\left[ \sqrt {k}\right] ^{\infty }_{K}$$
as before and this is infinite.
No doubt this needs tidying up for more rigorous expression but I think a version of that stands up.

I don't do that now as I see something better for the problem, but we and other contributions have not totally wasted our time. Still, you practically had this yourself before:
$$\sum ^{N}_{k=1}(\sqrt {1+k}-\sqrt {k}) = $$
$√2 - √1 + √3 - √2 + √4 - √3 +... + √(N+1) - √N$
$= √(N + 1) - 1$
which increases with $N$ without limit.