Limit and infinite sums. Finding $\lim_{x\rightarrow\infty}\sum^{\infty}_{k=1}\frac{1}{k^3 x-k^2}$

Question

Could anyone help me with this problem. Compute $$\lim_{x\rightarrow\infty}\sum^{\infty}_{k=1}\dfrac{1}{k^3 x-k^2}$$ I don't know how to change a limit and a sum. Could you help me with this problem then suggest me some materials about changing a limit and a sum. Thanks.

Answer 1

In your Task you have applied

(i) at first a sum

(ii) then the Limit.

When you Change a sum with the Limit it means that you will have

(i) at first the Limit

(ii) then the sum.

You will end up with a sum over $lim_{x \rightarrow \infty} f_x(k)$ for any Expression $f_x(k)$.

Answer 2

Squeezing is a fast and easy option. For any $x>1$ we have: $$0\leq \sum_{k\geq 1}\frac{1}{k^3 x-k^2}\leq \frac{1}{x-1}\sum_{k\geq 1}\frac{1}{k^3} = \frac{\zeta(3)}{x-1}$$ so the limit is zero.

Answer 3

One may notice that your initial series admits a closed form.

Step 1. For $x$ sufficiently great, by partial fraction decomposition, you have $$\frac{1}{k^3x-k^2}=-\frac{1}{k^2}+x\left(-\frac{1}{k}+\frac{1}{k-\frac 1x}\right) , \quad k\geq1.\tag1$$

Step 2. Recall that the digamma function admits the following series representation $$\begin{equation} \psi(x+1) = -\gamma + \sum_{k=1}^{\infty} \left( \frac{1}{k} - \frac{1}{x+k} \right), \quad x >-1, \tag2 \end{equation} $$ where $\gamma$ is the Euler-Mascheroni constant.

Step 3. Combining $(1)$ and $(2)$, you get $$\sum_{k=1}^{\infty}\frac{1}{k^3x-k^2}=-\frac{\pi^2}{6}-\gamma \:x - x \:\psi\left(1-\frac 1x\right)\tag3$$

Step 4. For $x$ sufficiently great, you have the following asymptotic expansion $$\psi\left(1-\frac 1x\right)=-\gamma -\frac{\pi^2}{6\:x}-\frac{\zeta (3)}{x^2}-\frac{\zeta (4)}{x^3} +\mathcal{O}\left(\frac1{x^4}\right)\tag4$$

Step 5. As $x \to \infty$, you get the following asymptotic expansion $$\sum_{k=1}^{\infty}\frac{1}{k^3x-k^2}=\frac{\zeta (3)}{x}+\frac{\pi^4}{90 \:x^2} +\mathcal{O}\left(\frac1{x^3}\right)\tag5$$

and your desired limit is equal to $0$.

Remark. One may directly obtain $(5)$, just by writing $$ \frac{1}{k^3x-k^2}=\frac{1}{x}\frac{1}{k^3\left(1-\frac 1{x k}\right)}=\frac 1{x \:k^3}+\frac 1{x^2 \:k^4}+\mathcal{O}\left(\frac1{x^3 \: k^5}\right) $$ and summing by the use of normal convergence on $x \in [a,+\infty)$, $a>1$.

Answer 4

Choosing $x>1$ and considering the uniform convergence, we have that

$$\lim_{x\rightarrow\infty}\sum^{\infty}_{k=1}\dfrac{1}{k^3 x-k^2}=\sum^{\infty}_{k=1}\lim_{x\rightarrow\infty}\dfrac{1}{k^3 x-k^2}=0$$