evaluate the limit

Vali

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\(\displaystyle a_{0},a_{1},...,a_{k}\) 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.
 

Jomo

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\(\displaystyle a_{0},a_{1},...,a_{k}\) 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?
 
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Vali

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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 \(\displaystyle L=a_{1}\cdot 0+a_{2}\cdot 0+...+a_{k}\cdot 0=0\)
 

Jomo

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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 + ... +ak = 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.
 
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pka

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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

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

Jomo

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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

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I'm 99% sure it's 0.
 

pka

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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:
\(\displaystyle \begin{align*}\left(\sqrt[3]{n+k}-\sqrt[3]{n}\right)&=\left(\dfrac{1}{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right) \end{align*}\)
\(\displaystyle L=\displaystyle\lim_{n\rightarrow \infty }\left(\dfrac{1}{\sqrt[3]{(n+k)^2}+\sqrt[3]{(n+1)}\sqrt[3]{n}+\sqrt[3]{n^2}}\right)=0\)
 
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