evaluate the limit

[Vali]

$\displaystyle a_{0},a_{1},...,a_{k}$ are real numbers and $\displaystyle a_{0}+a_{1}+...+a_{k}=0$
$\displaystyle L=\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$
$\displaystyle L=?$

I have no idea how to start, I think I should write difference of sqrt to find the limit.

 

[Steven G]

$\displaystyle a_{0},a_{1},...,a_{k}$ are real numbers and $\displaystyle a_{0}+a_{1}+...+a_{k}=0$
$\displaystyle L=\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$
$\displaystyle L=?$

I have no idea how to start, I think I should write difference of sqrt to find the limit.

What can you say about $\displaystyle L=\lim_{n\rightarrow \infty }(a_{r}\sqrt[3]{n+r})$ for any r > 0? That is, what, if anything, is the difference between these limits with different r?

 

[Vali]

I think there's no difference because everytime L tends to infinity
Meanwhile, I think I found a solution.

$\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}$ so $\displaystyle L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$


$\displaystyle L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$


Now I group the terms and I have


$\displaystyle L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$


I split in more limits, the constants $\displaystyle a_{0},a_{1},...,a_{k}$ go ahead the limits and the limits are 0 after some calculations.In the end $\displaystyle L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0$

 

[Steven G]

I think there's no difference because everytime L tends to infinity
Meanwhile, I think I found a solution.

$\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}$ so $\displaystyle L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$


$\displaystyle L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$


Now I group the terms and I have


$\displaystyle L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$ On one hand we get [tex}\infty-\infty[/tex]


I split in more limits, the constants $\displaystyle a_{0},a_{1},...,a_{k}$ go ahead the limits and the limits are 0 after some calculations.In the end $\displaystyle L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0$

So you are not using a₁ + a₂ + ... +a_k = 0. It could be possible that this is not needed. My main concern is that you think that inf - inf = 0. This is not true at all. Consider $\displaystyle L=\lim_{n\rightarrow \infty }[(n+k) - (n)]$. On one hand, you get $\displaystyle \infty - \infty$, while on the otherhand, n+k - n =k, so the $\displaystyle L=\lim_{n\rightarrow \infty }[(n+k) - (n)]$ = k, for some number k.

So in the end we see that some limits that give us $\displaystyle \infty - \infty$ can actually equal ANY NUMBER.

 

[pka]

I think there's no difference because everytime L tends to infinity
Meanwhile, I think I found a solution
$\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}$ so $\displaystyle L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$ (1)
$\displaystyle L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$ (2)
Now I group the terms and I have
$\displaystyle L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$ (3)

I think you, Vali, are correct in that $\displaystyle \large\color{blue}{(1)\to(2)\to(3)}$

Lets point out that:
\(\displaystyle \begin{align*}\left(\sqrt[3]{n+k}-\sqrt[3]{n}\right)&=\left(\sqrt[3]{n+k}-\sqrt[3]{n}\right)\left(\dfrac{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}{\sqrt[3]{(n+k)^2}\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right)\\&=\left(\dfrac{1}{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right) \end{align*}\)

 

[Vali]

So the limit is not 0 too ?
Because we have n at denominator

 

[Steven G]

So the limit is not 0 too ?
Because we have n at denominator

Why do you think that? For example what does 1/n approach as n approaches infinity?

 

[Vali]

I'm 99% sure it's 0.

 

[pka]

I think there's no difference because everytime L tends to infinity Meanwhile, I think I found a solution.
$\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}$ so $\displaystyle L=\lim_{n\rightarrow \infty }[(-a_{1}-a_{2}-...-a_{k})\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k}]$
$\displaystyle L=\lim_{n\rightarrow \infty }(-a_{1}\sqrt[3]{n}-a_{2}\sqrt[3]{n}-...-a_{k}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})$
Now I group the terms and I have
$\displaystyle L=\lim_{n\rightarrow \infty }[a_{1}(\sqrt[3]{n+1}-\sqrt[3]{n})+a_{2}(\sqrt[3]{n+2}-\sqrt[3]{n})+...+a_{k}(\sqrt[3]{n+k}-\sqrt[3]{n})]$
I split in more limits, the constants $\displaystyle a_{0},a_{1},...,a_{k}$ go ahead the limits and the limits are 0 after some calculations.In the end $\displaystyle L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0$

So the limit is not 0 too ?
Because we have n at denominator

I'm 99% sure it's 0.

Lets point out that:
\(\displaystyle \begin{align*}\left(\sqrt[3]{n+k}-\sqrt[3]{n}\right)&=\left(\dfrac{1}{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right) \end{align*}\)

$\displaystyle L=\displaystyle\lim_{n\rightarrow \infty }\left(\dfrac{1}{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right)=0$