-1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ did you try root test? $\endgroup$ – Alex Apr 7 '15 at 22:19
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

just use the root test to show divergence

$\endgroup$
0
$\begingroup$

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 $b_k=1$)

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ There is a $\sqrt k$ in the dinomiator so the equivalent would be correct if you multiply $b_k$ by $(\sqrt k)^k$ $\endgroup$ – Elaqqad Apr 7 '15 at 22:25
  • $\begingroup$ @Elaqqad I had corrected that, but was still missing a $e^{7/12}$ constant. $\endgroup$ – Clement C. Apr 7 '15 at 22:31
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.