Prove the divergence of $\sum_{k=1}^{\infty}{\frac{k}{\sqrt[3]{k+1}}}$ (Alternative proofs)

Question

I argued like that:$$ \lim\limits_{k\to\infty}\frac{k}{\sqrt[3]{k+1}}=\lim\limits_{k\to\infty}\frac{\sqrt[3]{k}\cdot \sqrt[3]{k^2}}{\sqrt[3]{k}\cdot \sqrt[3]{1+\frac{1}{k}}}=\lim\limits_{k\to\infty}\frac{\sqrt[3]{k^2}}{\sqrt[3]{1+\frac{1}{k}}}\to \infty$$ Since the sequence isn't a null sequence (tending to zero), the series diverges.

Could you also show this with the ratio test, root test, direct comparison test?

Answer 1

If I understand correctly the terms, the "ratio test" and the "root test" are the computation of the limits of respectively $\frac{u_{n+1}}{u_n}$ and $\sqrt[n]{u_n}$.

As the general term of the series is proportional to a Riemann series term, you can't apply those tests : you will always find a limit $1$ : $$ \frac{(n+1)^\alpha}{n^\alpha} = (1+\frac1n)^\alpha \xrightarrow[n\to\infty]{}1$$ and $$\sqrt[n]{n^\alpha}=n^{\alpha/n} = e^{\frac\alpha n\ln n}\xrightarrow[n\to\infty]{}1$$ Those tests are for comparison with a rapidly growing or collapsing general term (their demonstrations compares the term to one of a geometric series).

Answer 2

Use that $$\frac{k}{\sqrt[3]{k+1}}\ge \frac{1}{k}$$ if $$k^6\geq k+1$$

Answer 3

$m:=k+1$, consider

$\displaystyle { \sum_{m=2}^{\infty}} \dfrac{m-1}{m^{1/3}}$.

$a_m:=m^{2/3}-m^{-1/3}$.

$\lim_{m \rightarrow \infty} a_m \not =0$.