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

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

$f'$ 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

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