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Suppose for each $n$, $a_{n+1} - a_{n} = \alpha > 0$ with $\alpha$ independent of $n$ and $a_1 > 0$. Show that the series $\sum_{n=1}^{\infty}\frac{1}{a_n}$ diverges.

My solution is as such:

$$a_{n+1} - a_{n} = \alpha$$ $$\implies a_{n+1} = a_{n} +\alpha = a_{n-1}+ 2\alpha = a_{1} + n\alpha$$

Hence, $$\sum_{n=1}^{k}\frac{1}{a_n} =\frac{1}{a_1}+\frac{1}{a_2}+...\frac{1}{a_k} \\= \frac{1}{a_1}+\frac{1}{a_1+\alpha}+\frac{1}{a_1+2\alpha}...+\frac{1}{a_1+k\alpha} \\\geq \frac{1}{a_1}+\frac{1}{a_1+\alpha}+\frac{1}{2(a_1+\alpha)}...+\frac{1}{k(a_1+\alpha)}\\=\frac{1}{a_1} +\frac{1}{a_1+\alpha}(1+\frac{1}{2}+...+\frac{1}{k}) \\ = \frac{1}{a_1} +\frac{1}{a_1+\alpha}(\frac{k(k+1)}{2})\\= \frac{1}{a_1} +\frac{k(k+1)}{2(a_1+\alpha)}$$

Since $$\frac{1}{a_1} +\frac{k(k+1)}{2(a_1+\alpha)} \text{diverges.}$$

By Comparison Test, $$\sum_{n=1}^{k}\frac{1}{a_n} \text{diverges.}$$

Therefore, $$\sum_{n=1}^{\infty}\frac{1}{a_n} \text{diverges.}$$

Is it correct?

Hamid Shafie Asl
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user10024395
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    You have committed an error in assuming $1 + 1/2 + ... + 1/n = n(n + 1) / 2$ as this does not hold true. The left hand side is the harmonic series and the right hand side the formula for an arithmetic series. It is however true (and "well known") that $1 + 1/2 + ... + 1/n + ...$ is divergent so the argument could be reworked with relative ease. – Squid Mar 18 '14 at 15:36

3 Answers3

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Consider $1+ \frac{1}{2} + \cdots + \frac{1}{n}=\sum_{k=1}^n\frac{1}{k}$. Then $\sum_{k=1}^\infty\frac{1}{k}=\lim_{n\to\infty}\sum_{k=1}^n\frac{1}{k}$ itself is diverges. So $\frac{1}{a_1+\alpha}\sum_{k=1}^\infty\frac{1}{k}$ is diverges and consequently $\frac{1}{a_1}+\frac{1}{a_1+\alpha}\sum_{k=1}^\infty\frac{1}{k}$ is diverges. So yours is diverges.

Hamid Shafie Asl
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This passage between the two lines is not correct:

$1+ \frac{1}{2} + \cdots + \frac{1}{n} \neq \frac{n(n+1)}{2}$

The true formula is : $1+ 2 + \cdots + n = \frac{n(n+1)}{2}$

For solving the exercice, let's remark that for $n \geq 1$, we have:

$a_n= (n-1) \alpha +a_1 \underset{+\infty}{\sim} n \alpha$

Pi89
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Your solution is good up until $$ \frac{1}{a_1} +\frac{1}{a_1+\alpha} \left(1+\frac{1}{2}+...+\frac{1}{n}\right) = \frac{1}{a_1} +\frac{1}{a_1+\alpha}\left(\frac{n(n+1)}{2}\right) $$

This is an error; you are thinking of the formula $1 + 2 + \cdots + n = \frac{n(n+1)}{2}$, which does not apply here. Actually, $1+\frac{1}{2}+...+\frac{1}{n}$ is the $n$th Harmonic Number and has no known closed form.

It is a partial sum of the Harmonic Series, so if you know already that the harmonic series diverges, you can just use that directly to show that your series diverges. If you don't know the harmonic series diverges, there are a number of ways to prove it, many of which have been discussed on this site before (1, 2, 3).

Community
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Caleb Stanford
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