Question

Determine whether the sequence an=12n3+22n3+⋯+n2n3 converges or diverges. if it converges, find the limit. note, ∑i=1ki2=k(k+1)(2k+1)6

Answer 1

I guess the sequence is

$a_n=\dfrac{1^2}{n^3}+\dfrac{2^2}{n^3}+\cdots\dfrac{n^2}{n^3}$

which we can write as

$a_n=\displaystyle\frac1{n^3}\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6n^3}=\dfrac{2n^2+3n+1}{6n^2}$

$a_n$ converges if it is bounded and monotonic. Consider the function,

$f(x)=\dfrac{2x^2+3x+1}{6x^2}=\dfrac13+\dfrac1{2x}+\dfrac1{6x^2}$

which has derivative

$f'(x)=-\dfrac1{2x^2}-\dfrac1{3x^3}$

$f'<0$ for all $x>0$, so $f(x)$ is monotonically decreasing on $(0,\infty)$, and as $x\to\infty$ we have

$\displaystyle\lim_{x\to\infty}f(x)=\dfrac13$

So we know that $a_n$ is monotonically decreasing and bounded below by $\dfrac13$.

To find the limit, we can also write

$a_n=\dfrac{2+\frac3n+\frac1{n^2}}6$

As $n\to\infty$, the rational terms vanish and we're left with

$\displaystyle\lim_{n\to\infty}\frac26=\frac13$

Answer 2

The converges and limit will be $\dfrac{1}{3}\ and\ \dfrac{1}{3}$

Sequence and series

It is the sum of the sequence of terms.

Given

$\rm a_{n} = \dfrac{1^{2} }{n^{3} } + \dfrac{2^{2} }{n^{3} } + ........\ +\dfrac{n^{2} }{n^{3} }$

How to calculate converges and limit?

$a_{n} = \dfrac{1}{n^{3} } \sum_{k=1}^{n} k^{2}$

$a_{n}$ converges if it is bounded and monotonic

$f(x)=\dfrac{2x^{2} +3x+1}{6x^{2} } = \dfrac{1}{3} + \dfrac{1}{2x} + \dfrac{1}{6x^{2} }$

And it's derivative

$f'(x) = - \dfrac{1}{2x^{2} } - \dfrac{1}{3x^{3} }$

And   $\displaystyle \lim_{ x\to \infty } f(x)=\dfrac{1}{3}$

To find the limit,

$a_{n} = \dfrac{2+ \frac{3}{n} + \frac{1}{n^{2} } }{6}$

As $n\to \infty$

Then

$a_{n} = \dfrac{1}{3}$

Thus, the converges and limit will be $\dfrac{1}{3}\ and\ \dfrac{1}{3}$

More about the Sequence and series link is given below.

https://brainly.com/question/8195467