Question

The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit in part (a). 13 23 33 n3

Answer 1

Answer:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$

Step-by-step explanation:

Assuming this complete problem: "The following formula for the sum of the cubes of the first n integers is proved in Appendix E. Use it to evaluate the limit . 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2"

We have the following formula in order to find the sum of cubes:

$\lim_{n\to\infty} \sum_{n=1}^{\infty} i^3$

We can express this formula like this:

$\lim_{n\to\infty} \sum_{n=1}^{\infty}i^3 =\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

And using this property we need to proof that: 1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

$\lim_{n\to\infty} [\frac{n(n+1)}{2}]^2$

If we operate and we take out the 1/4 as a factor we got this:

$\lim_{n\to\infty} \frac{n^2(n+1)^2}{n^4}$

We can cancel $n^2$ and we got

$\lim_{n\to\infty} \frac{(n+1)^2}{n^2}$

We can reorder the terms like this:

$\lim_{n\to\infty} (\frac{n+1}{n})^2$

We can do some algebra and we got:

$\lim_{n\to\infty} (1+\frac{1}{n})^2$

We can solve the square and we got:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2})$

And when we apply the limit we got that:

$\lim_{n\to\infty} (1+ \frac{2}{n} +\frac{1}{n^2}) =1$