Answer to Question #265711 in Calculus for Sayem

Question

Determine whether the following series converge, converge absolutely, converge conditionally, or diverge.

1. $\displaystyle\sum_{k=1}^ ∞$ 3^k/k²*2^k
2. $\displaystyle\sum_{k=1}^ ∞$ (-1)^k+1 / ( √(k+1) -√k)
3. $\displaystyle\sum_{k=1}^ ∞$ (-1)^k+1 [√(k+1) -√k)]

Answer 1

Let us determine whether the following series converge, converge absolutely, converge conditionally, or diverge.

1) $\displaystyle\sum_{k=1}^{\infty}\frac{3^k}{k^22^k}$ 

Let us use D'Alembert's Ratio Test.

Since $\lim\limits_{n\to\infty}(\frac{3^{k+1}}{(k+1)^22^{k+1}}:\frac{3^k}{k^22^k}) =\lim\limits_{n\to\infty}\frac{3k^2}{2(k+1)^2}=\lim\limits_{n\to\infty}\frac{3k^2}{2(k^2+2k+1)}=\frac{3}2>1,$

we conclude that this series diverge.

2) Since

$\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{ \sqrt{k+1} -\sqrt{k}} =\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(\sqrt{k+1} +\sqrt{k})}{ (\sqrt{k+1} -\sqrt{k})(\sqrt{k+1} -\sqrt{k})} =\displaystyle\sum_{k=1}^{\infty}\frac{(-1)^{k+1}(\sqrt{k+1} +\sqrt{k})}{ ((k+1)-k)} =\displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}(\sqrt{k+1} +\sqrt{k}),$ 

and $\lim\limits_{k\to\infty}(\sqrt{k+1} +\sqrt{k})=\infty,$

we conclude that the necessary condition of covergence is not met, and hence this series is divergent.

3) $\displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}( \sqrt{k+1} -\sqrt{k})$

Consider the series $\displaystyle\sum_{k=1}^{\infty}|(-1)^{k+1}( \sqrt{k+1} -\sqrt{k})|=\displaystyle\sum_{k=1}^{\infty}(\sqrt{k+1} -\sqrt{k}).$

Since $A_n=(\sqrt{2} -1)+(\sqrt{3} -\sqrt{2})+(\sqrt{4} -\sqrt{3})+\ldots +(\sqrt{n+1} -\sqrt{n})=\sqrt{n+1} -1,$

and $\lim\limits_{n\to\infty}A_n=\lim\limits_{n\to\infty}(\sqrt{n+1} -1)=\infty,$ we conclude that the series $\displaystyle\sum_{k=1}^{\infty}(-1)^{k+1}( \sqrt{k+1} -\sqrt{k})$ diverge absolutely.