Divergence for Series with $|a_n| \ge b_n \ge 0$

Question

first of all i hope you get my point since English is not my native.

I know the series $$ y = \sum_{k=1}^\infty \frac{1}{k} $$ is divergent.

Now I want to check if the series $$x=\sum_{k=1}^\infty \frac{1}{2(k+1)} $$ is divergent too.

I "have" to use the criteria with $$ |a_k| \ge d_k \ge 0 $$ in Germany it is called Minorante.

So when $\sum d_k$ is divergent obviously $\sum a_k$ is divergent too. In my case with an indexshift we can put $$ \sum_{k=1}^\infty \frac{1}{2(k+1)} $$ into $$ 1/2 * \sum_{k=2}^\infty 1/k $$

We see when we shift the Index in the other formula $$ y = \sum_{k=1}^\infty 1/k $$ to $$ y = \sum_{k=2}^\infty 1/k $$ we have nearly the same expression only that we have a *1/2 on $x_k$.

My question is, the definition is $$ |a_k| \ge d_k \ge 0. $$ For our case $$ x \ge y \ge 0 $$ Why is our $x_k = a_k $ and our $y_k = d_k $?

Because $ y = \sum_{k=2}^\infty 1/k $ is $\gt$ than $ x = 1/2*\sum_{k=2}^\infty 1/k $

Or do we ignore the 1/2*? Then it would be at least equal and the definition would be okay.

My solution from the university says: We know $y = \sum_{k=2}^\infty 1/k$ is divergent so we can conclude with the defintion $|a_k| \ge d_k \ge 0$ that $x = \sum_{k=1}^\infty 1/(2(k+1))$ is divergent too.

Answer 1

Hint

• $\displaystyle\frac{1}{2(k+1)}=\frac{1}{2k+2}\ge\frac{1}{4k}, \forall k\in \mathbb{N}.$
• The series $\displaystyle\sum_{k=1}^{\infty}\frac{1}{4k}=\frac 14 \sum_{k=1}^{\infty}\frac{1}{k}$ is divergent.