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

\(\displaystyle L=?\)

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

- Thread starter Vali
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\(\displaystyle L=\lim_{n\rightarrow \infty }(a_{0}\sqrt[3]{n}+a_{1}\sqrt[3]{n+1}+...+a_{k}\sqrt[3]{n+k})\)

\(\displaystyle L=?\)

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

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

\(\displaystyle L=?\)

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

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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 you are not using a[SUB]1 [/SUB]+ a[SUB]2[/SUB] + ... +a[SUB]k[/SUB] = 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 \(\displaystyle L=\lim_{n\rightarrow \infty }[(n+k) - ]\). On one hand, you get \(\displaystyle \infty - \infty\), while on the otherhand, n+k - n =k, so the \(\displaystyle L=\lim_{n\rightarrow \infty }[(n+k) - ]\) = k, for some number k.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 in the end we see that some limits that give us \(\displaystyle \infty - \infty\) can actually equal ANY NUMBER.

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

Lets point out that:

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

Why do you think that? For example what does 1/n approach as n approaches infinity?So the limit is not 0 too ?

Because we have n at denominator

\(\displaystyle a_{0}=-a_{1}-a_{2}-...-a_{k}\) 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.

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

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