Csdrhrt

I am a bit perplexed in trying to find values $a,b,c$ so that the approximation is as precise as possible:

$$sum_{k=n}^{infty}frac{(ln(k))^{2}}{k^{3}} approx frac{1}{n^{2}}[a(ln (n))^{2}+b ln(n) + c]$$

I can see from Wolfram that $ln(x)$ could be written in many different ways and that the sum starting from $1$ can be written with Riemann zeta function here. They are probably not something I am looking here. The decreasing exponent hints me that perhaps the term in the sum can be approximated with its derivations, the first derivate here and the second here but when I try to approximate something goes wrong, perhaps wrong premise. Let $z=frac{ln(k)^{2}}{k^{3}}$ then

$$sum_{k=n}^{infty} z approx z' z-z' z^3/2!$$

by the Taylor approximation for the odd function ($ln(x)$ is odd, $x^3$ is odd and the oddity is preserved after the operations, maybe wrong, so $z$ is odd like $sin(x)$), I get:

$$sum_{k=n}^{infty} z approx (frac{1}{x^3}(2ln(x) - 3(ln(x))^2)) -frac{1}{x^2}(6 ln(x)^2-7 ln(x)+1)$$

something wrong because the first term has $x^3$ instead of $x^2$. I am sorry if this hard to read but stuck here. So is it the correct way to approximate the sum? Is it really with Riemann zeta function or does Taylor work here, as I am trying to do above?

I labeled the question with parity because I think it is central here. I feel I may be misunderstanding how the oddity works over different operations, hopefully shown in my vague explanations, and hence the error.

You can try using elementary estimates (upper and lower bounds) with integrals (see the end of the answer), or can use the more general and quite useful Euler Mclaurin Summation Formula, which gives us

$$sum_{k=1}^{n} frac{log^2 k}{k^3} = int_{1}^{n} frac{log^2 x}{x^3} text{d}x + frac{log ^2 n}{2n^3} + C + mathcal{O}(frac{1}{n^{3+e}})$$

where $displaystyle e > 0$.

Now $$int_{1}^{n} frac{log^2 x}{x^3} text{d}x = - frac{2log^2 n + 2 log n + 1}{4n^2} + 1/4$$

Thus

$$sum_{k=1}^{n} frac{log^2 k}{k^3} = K - frac{2log^2 n + 2 log n + 1}{4n^2} + frac{log^2 n}{2n^3} + mathcal{O}left(frac{1}{n^{3+e}}right)$$

Since the LHS converges we have that

$$sum_{k=1}^{n} frac{log^2 k}{k^3} = sum_{k=1}^{infty}frac{log^2 k}{k^3} - frac{2log^2 n + 2 log n + 1}{4n^2} + frac{log^2 n}{2n^3} + mathcal{O}left(frac{1}{n^{3+e}}right)$$

Thus

$$sum_{k=n+1}^{infty}frac{log^2 k}{k^3} = frac{2log^2 n + 2 log n + 1}{4n^2} - frac{log^2 n}{2n^3} + mathcal{O}left(frac{1}{n^{3+e}}right)$$

And so

$$sum_{k=n}^{infty}frac{log^2 k}{k^3} = frac{2log^2 n + 2 log n + 1}{4n^2} + frac{log^2 n}{2n^3} + mathcal{O}left(frac{1}{n^{3+e}}right)$$

Note that the summation formula allows us to figure out the exact terms which make up the $displaystyle mathcal{O}left(frac{1}{n^{3+e}}right)$ and so we can make the formula more accurate by including as many lower order terms we need.

A more elementary way:

If $displaystyle f(x) ge 0$ and is monotonically decreasing then we have that

$$int_{n}^{M+1} f(x) text{d}x le sum_{k=n}^{M} f(k) le f(n) + int_{n}^{M} f(x) text{d}x$$

which can be seen by bounding the area under the curve by rectangles of width $1$.

For $displaystyle f(n) = frac{log ^2 n}{n^3}$, we can take limit as $displaystyle M to infty$ to get the asymptotic formula you need.

Invoke an Integral test for convergence
then two Integration by parts give you (up to a mistake in my calculus)

$a=1/2$, $b=1/2$, and $c=1/4$.