Convergence of the series $ \sum\frac{1}{a_na_{n+1}} $ if the sequence $(a_n)$ is arithmetic

Question

We know that $(a_n)$ is arithmetic progression and I need to decide if $$ \sum_{n=1}^{\infty} \frac{1}{a_n \cdot a_{n+1}} $$ convergences. Firstly I check necessary condition: $ \frac{1}{a_n \cdot a_{n+1}} \rightarrow 0 $ It is true because $$ a_n \rightarrow +/- \infty $$ because $ a_n = a + (n-1)r $

Ok, now I should check convergence: Ratio test: $$ \frac{u_{n+1}}{u_n} = \frac{a_n(a_n+r)}{(a_n+r)(a_n+2r)} \rightarrow 1 $$ So it gives me nothing. Root tests: $$ \sqrt[n]{\frac{1}{a_n \cdot a_{n+1}}} \rightarrow 1 $$ so it fails too... Have somebody any idea how can I check this?

Answer 1

First note that $$\frac{1}{a_na_{n+1}}=\frac{1}{a_{n+1}-a_n}\cdot \left(\frac{1}{a_{n}}-\frac{1}{a_{n+1}}\right)$$

Then since $a_n$ is arrithematic, $a_{n+1}-a_n=a_2-a_1=d.$ So you get:

$$\sum_{n=1}^{N}\frac{1}{a_na_{n+1}}=\frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{N+1}}\right)$$

Which means the series converges to...

We have to assume the $a_i\neq 0$ and that $d\neq 0.$ If $d=0$ then all the $a_i$ are equal, and the sum diverges.

Answer 2

If $a_n$ is an arithmetic progression, then $a_n$ grows with order $n$; therefore, $a_n \cdot a_{n + 1}$ grows at least as quickly as $n^2$, and convergence follows from a comparison test. Note that this convergence is too slow to be detected by either ratio test or root test.

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To make this a bit more precise, if $a_n$ is a term of an arithmetic progression we can write $a_n =cn + b$ for constants $c$ and $b$. Then

$$\frac{1}{a_n \cdot a_{n + 1}} = \frac{1}{c^2 n^2 + \text{ lower order}}$$

and the comparison is quick from here.