TransWikia.com

Is this a correct way to show that $sum_{k=2}^{infty} frac {1}{sqrt k+k(-1)^k}$ converges?

Mathematics Asked by Labbsserts on February 9, 2021

Show that $sum_{k=2}^{infty} frac {1}{sqrt k+k(-1)^k}$ converges or diverges. My attempt is ot first rewrite and see if the sequence is convergent according to the Leibniz test:

$frac {1}{sqrt k+k(-1)^k}= frac{(-1)^k}{sqrt k} frac{1}{sqrt k + (-1)^k} lefrac{(-1)^k}{sqrt k} frac{1}{sqrt k -1}$

Now see if it holds that:

  • $a_{k+1} le a_k$
  • $lim_{k rightarrow infty} a_k = 0$

The first criterion holds because the denominator is an increasing functions and thus an increasing sequence:
$f(x) = x-sqrt x implies f'(x)=1-frac{1}{2sqrt x} ge 0 forall x>2$

The second criterion also holds because: $lim_{k rightarrow infty} frac{1}{sqrt k} frac{1}{sqrt k-1} rightarrow 0 * 0=0 $. So according to the Leibniz test the sum converges

One Answer

It is true that $a_k=frac1{sqrt{k}+(-1)^kk}$ is less than $b_k=frac{(-1)^k}{sqrt{k}(sqrt{k}-1)}$. But each of these sequences is alternating in sign. And even if $sum_{k=2}^infty b_k$ converges, the dominance of $b_k$ over $a_k$ does not guarantee the convergence of $sum_{k=2}^infty a_k$.

However, we can proceed as follows. Note that we can write

$$begin{align} frac1{sqrt{k}+(-1)^kk}&=frac{sqrt{k}-(-1)^kk}{k(1-k)}\\ &=frac{1}{sqrt{k}(1-k)}-frac{(-1)^k}{1-k} end{align}$$

Hence, we can write

$$begin{align} sum_{k=2}^K frac1{sqrt{k}+(-1)^kk}&=sum_{k=2}^K frac{1}{sqrt{k}(1-k)}-sum_{k=2}^Kfrac{(-1)^k}{1-k}tag1 end{align}$$

The first sum on the right-hand side of $(1)$ converges as $Kto infty$ by comparison to $sum_{k=1}^infty frac1{k^{3/2}}$ and the second sum converges as guaranteed by Leibniz's Test.

Answered by Mark Viola on February 9, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP