Help with proof that $\sum_{n \in \Bbb{N}} \frac{1}{an + b}$ also diverges?

Question

We know that $\sum_{n \in \Bbb{N}} \frac{1}{n}$ diverges. So it seems likely that $\sum_{n \in \Bbb{N}} \frac{1}{a n + b}$ will for any real $a, b$. I'm having trouble proving it just for the $\frac{1}{n + c}$ form special case. Any hints?

Answer 1

Let $a,b \in \mathbb{R}$ and let $a \neq 0$. Since $$ \frac{1}{an+b} \sim \frac{1}{an} $$ as $n$ grows, that is to say, $$ \frac{n}{an+b} \to \frac{1}{a} $$ as $n$ grows, so by the limit comparison test the series $\sum 1/(an+b)$ diverges.

Answer 2

Hint(s):

$$\sum_{n \in \mathbb N} \frac{1}{n+c}=\sum_{n \in \mathbb N} \frac{1}{n}- \sum_{k=1}^c \frac {1}{k}$$

And $\frac{1}{an+b}$ can be rewritten as $\frac{1}{a} \cdot \frac{1}{n+\frac{b}{a}}$. Now let $\frac{b}{a}:=c$.

Answer 3

First, since $\sum_{n=1}^M \frac1{n} \to \infty $ as $M \to \infty $, then, for any $m > 0$, $\sum_{n=m}^M \frac1{n} \to \infty $ as $M \to \infty $. Do you see why?

Now, let's look at the partial sums of the problem's series.

$\sum_{n=1}^M \frac1{an+b} =a\sum_{n=1}^M \frac1{n+b/a} $. Let $k$ be an integer such that $k \ge b/a$. Then $\frac1{n+b/a} \ge \frac1{n+k} $, so $\sum_{n=1}^M \frac1{n+b/a} \ge \sum_{n=1}^M \frac1{n+k} = \sum_{n=k+1}^{M+k} \frac1{n} $.

By the comment above, $\sum_{n=k+1}^{M+k} \frac1{n} \to \infty $ as $M \to \infty $.

Therefore $\sum_{n=1}^M \frac1{an+b} \to \infty $ as $M \to \infty $.

Answer 4

It very simple to proof that if $(*)\frac{|u_n|}{|v_n|}\to L\neq 0,\infty$ so $$\sum_n |u_n|<\infty \qquad \iff\qquad \sum_n |v_n|<\infty$$ in fact $(*)$ implies that : for all $L>\epsilon>0$ it exist $N_0\in\mathbb{N}$ such that $\forall n\geq N_0$ $\left| \frac{|u_n|}{|v_n|}- L\right|<\epsilon$ and this implies that for $N_0\in\mathbb{N}$ $$ (L-\epsilon)|v_n|<|u_n|<(L+\epsilon)|v_n| $$ So $$ (L-\epsilon)\sum_{n\geq N_0}|v_n|\leq \sum_{n\geq N_0} |u_n|\leq (L+\epsilon)\sum_{n\geq N_0}|v_n| $$ and this proof our result.

Now we applies this result to your case : where $v_n=n$ and $u_n=an+b$.

So $\frac{|u_n|}{|v_n|}\to a$

if $a>0$ so $\exists N \in \mathbb{N}$ such that $v_n\geq0$ for all $n\geq N$ and then $\sum_n v_n =\infty$.

If $a<0$ so $\exists N' \in \mathbb{N}$ such that $-v_n\geq0$ for all $n\geq N'$ and then $-\sum_n v_n =\infty$.

In booth cases $\sum_n\frac{1}{an+b}$ diverge for all $a,b\in\mathbb{R}$

Answer 5

Here's a quick one. $an + b \leq an + bn = (a + b) n$, so \begin{align*} \sum_{n = 1}^{\infty} \frac{1}{an + b} & \geq \sum_{n = 1}^{\infty} \frac{1}{an + bn} \\ & = \frac{1}{a + b} \sum_{n = 1}^{\infty} \frac{1}{n} \\ & = \frac{+ \infty}{a + b} \\ & = + \infty . \end{align*}