# check if the series $\sum_{k=1}^{\infty}\left(\frac{3k+1}{\sqrt{2k-1}}\right)^k$ converges

Check if the following series is convergent or divergent:
$$\sum_{k=1}^{\infty}\left(\frac{3k+1}{\sqrt{2k-1}}\right)^k$$ This series should be divergent too, one way to see this is the comparison test. Here I need a divergent series $\sum\limits_{k=1}^{\infty}b_k$ such that $0\le b_k \le \left(\frac{3k+1}{\sqrt{2k-1}}\right)^k$ for all sufficient large natural k, too. But I don't know what $b_k$ could be. Or is there an easier way to prove the divergence of this series?

• did you try root test? – Alex Apr 7 '15 at 22:19

Inside the parentheses the expression $\to \infty.$ If that wasn't impressive enough, we then raise to the $k$th power. So the $k$th term of the series blasts off to $\infty,$ and since $\infty\ne 0,$ the series diverges, big time.

just use the root test to show divergence

You can also see that $(2k-1)^2=(2k-1)(2k-1)\leq(3k+1)^2$ and take: $$b_k=(\sqrt{2k-1})^k$$

and the series is clearly divergent (or you kan just take $b_k=1$)

The general term $a_k \stackrel{\rm def}{=} \left(\frac{3k+1}{\sqrt{2k-1}}\right)^k$ is equivalent, when $k\to\infty$, to $$b_k \stackrel{\rm def}{=} e^{7/12}k^{k/2}\left(\frac{3}{\sqrt{2}}\right)^k$$ and since the (positive) series $\sum_k b_k$ diverges, so does $\sum_k a_k$ by comparison.

• There is a $\sqrt k$ in the dinomiator so the equivalent would be correct if you multiply $b_k$ by $(\sqrt k)^k$ – Elaqqad Apr 7 '15 at 22:25
• @Elaqqad I had corrected that, but was still missing a $e^{7/12}$ constant. – Clement C. Apr 7 '15 at 22:31

$(2k-1)^2\leq (3k+1)^2$ and thus $\infty =\sum_{k=1}^{\infty} 1 \leq \sum_{k=1}^{\infty}\left(\frac{3k+1}{\sqrt{2k-1}}\right)^k$

Call $a_k:=\frac{3k+1}{\sqrt{2k-1}}$. Clearly $a_k\to+\infty$ and so does $a_k^{k}$. Thus the series diverges.

Let

$a_ k = \left(\frac{3k+1}{\sqrt{2k-1}}\right)^k$

We have

$\ell = \lim\limits_{k \to \infty} \sqrt[k]{|a_k|} \Rightarrow \ell = \lim\limits_{k \to \infty} \sqrt[k]{\left|\left(\frac{3k+1}{\sqrt{2k-1}}\right)^k\right|} \Rightarrow \ell = \lim\limits_{k \to \infty}\underbrace{\left|\frac{3k+1}{\sqrt{2k-1}}\right|}_{> \ 0} \Rightarrow \ell = \lim\limits_{k \to \infty}\frac{3k+1}{\sqrt{2k-1}} \Rightarrow \boxed{\ell = \infty}$

Since $\ell > 1$, by the root test, the series diverges.

PS: $\sum\limits_{k = 0 }^{\infty} \left(\frac{3k+1}{\sqrt{2k-1}}\right)^{-k} \cong 1.3243840336\ldots$