Prove that $(a_n)_{n\geq1}$ is convergent, given that: $a_n=\frac{1}{3}+\frac{1}{4\sqrt[m]{2}}+...+\frac{1}{(n+2)\sqrt[m]{n}}$

Question

Considering $m\geq 2, m\in\mathbb{N}$, prove that the sequence $(a_n)_{n\geq1}$ is convergent, given that:
$a_n=\frac{1}{3}+\frac{1}{4\sqrt[m]{2}}+...+\frac{1}{(n+2)\sqrt[m]{n}}$
I tried to prove that $(a_n)$ is convergent by using the Monotone convergence theorem, because $a_n>0$ and $a_{n+1}-a_n > 0$ which implies $a_n$- increasing sequence.
But i can't find its boundary in order to say that its supremum is the limit.
Maybe my method isn't the right one, but i couldn't think of anything else. Any ideas for finding that boundary? Thank you in advance!

Answer 1

Compare the series with $\sum_{n=1}^{\infty} \frac{1}{n \sqrt[m]{n}} $. This series is convergent because $\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$ is convergent iff $\alpha > 1$, in this case $\alpha = 1+ \frac 1 m > 1$. Taking the limit

$$\lim_{n \to \infty} \frac{\frac{1}{(n+2) \sqrt[m]{n}}}{\frac{1}{n \sqrt[m]{n}}}= 1 $$

So then your series is also convergent by the comparison criterion by step to the limit.

Answer 2

$a_n=\sum_{n=1}^{\infty}\frac{1}{(n+2)\sqrt[m]{n}}<\sum_{n=1}^{\infty}\frac{1}{n\sqrt[m]{n}}=\sum_{n=1}^{\infty}\frac{1}{n\cdot n^{1/m}}=\sum_{n=1}^{\infty}\frac{1}{n^{1+1/m}}$

A series converges $\sum_{x=a}^{x=b} f(x)\iff \int^b_af(x)dx $ is finite

So $\int^{\infty}_1n^{-(1+1/m)}dn=|^{\infty}_1-mn^{-1/m}=m$

$\sum_{n=1}^{\infty}\frac{1}{n^{1+1/m}}$ converges