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I want to determine whether the series below is absolutely convergent, conditionally convergent or divergent. $\sum_{n=1}^{+\infty}c_n\text{, where }c_n =\begin{cases} -\frac{1}{n} \text{, if $\frac{1}{4}n$ is an integer} \\ \frac{1}{n^2} \text{, if $\frac{1}{4}n$ is not an integer}\end{cases}$

The terms $c_n$ of this series, for some initial values of $n$, are as follows:

$\begin{array}{c|c|c|} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline c_n & \frac{1}{1^2} & \frac{1}{2^2} & \frac{1}{3^2} & -\frac{1}{4} & \frac{1}{5^2} & \frac{1}{6^2} & \frac{1}{7^2} & -\frac{1}{8} \\ \hline \end{array}$

This series appears to be divergent, and I will make an attempt to show it.

First (and probably wrong) attempt

(Edit) NOTE: I think this attempt is wrong, because, when the series is divergent, I can't necessarily group the terms arbitrarily to form a new infinite series.

See second attempt below.

I will use the following reasoning: I can group the terms into groups of four to form a new infinite series which is equivalent to the previous one, as follows:

$\left(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4}\right) + \left(\frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} - \frac{1}{8}\right)+\cdots$

The $N$th partial sum of the series formed as such can be represented as:

$\sum_{i=1}^{N}\left(\frac{1}{(4n-3)^2}+\frac{1}{(4n-2)^2}+\frac{1}{(4n-1)^2}-\frac{1}{4n}\right)\ \ \ \ (*)$

The partial sum above can be rewritten as a sum of two partial sums:

$\sum_{i=1}^{N}\left(\frac{1}{(4n-3)^2}+\frac{1}{(4n-2)^2}+\frac{1}{(4n-1)^2}\right)+\sum_{i=1}^{N}\left(-\frac{1}{4n}\right) \ \ (**)$

If we take the limit of the above expression as $N\to+\infty$, we see that the sum in the left converges, but the sum in the right diverges. Therefore, for $N\to+\infty$, the sum is divergent.

Also, since the expression marked above with one asterisk can be seen as the term-by-term sum of the two series in the expression above marked with two asterisks, I can also use the fact that if we sum a convergent infinite series and a divergent infinite series term-by-term, the resulting series is divergent.

Update: Second attempt

I will try to make use of the following theorem:

Theorem: Let $\{s_n\}$ be the sequence of partial sums of a given convergent series $\sum_{n=1}^{+\infty}u_n$. Then, for any $\epsilon>0$, there is a number $N$ such that:

$|s_R - s_T| < \epsilon \text{ whenever } R>N \text{ and } T>N$

Below, I will assume that the series is convergent and try to apply the above theorem, to show by contradiction that the series is divergent. Thus, I will try to find a positive lower bound for $|s_{8n}-s_{4n}|$, which would show that the series diverges.

The expression for $s_{8n}-s_{4n}$ is:

$s_{8n}-s_{4n} = \left(\frac{1}{(4n+1)^2} + \frac{1}{(4n+2)^2} + \frac{1}{(4n+3)^2} - \frac{1}{4n+4}\right)+\left(\frac{1}{(4n+5)^2} + \frac{1}{(4n+6)^2} + \frac{1}{(4n+7)^2} - \frac{1}{4n+8}\right)+\cdots+\left(\frac{1}{(8n-3)^2} + \frac{1}{(8n-2)^2} + \frac{1}{(8n-1)^2} - \frac{1}{8n}\right)$

The sum above has $4n$ terms. Separating the positive terms from the negative terms, and forming two groups (one with all the positive terms, and other one with all the negative terms), we get:

$s_{8n}-s_{4n} = \left[\left(\frac{1}{(4n+1)^2} + \frac{1}{(4n+2)^2} + \frac{1}{(4n+3)^2}\right)+\left(\frac{1}{(4n+5)^2} + \frac{1}{(4n+6)^2} + \frac{1}{(4n+7)^2}\right)+\cdots+\left(\frac{1}{(8n-3)^2} + \frac{1}{(8n-2)^2} + \frac{1}{(8n-1)^2}\right)\right]-\left(\frac{1}{4n+4}+\frac{1}{4n+8}+\cdots+\frac{1}{8n}\right)$

For simplicity, I will rewrite the equation above as follows:

$s_{8n}-s_{4n} = f(n) - g(n)$

where

$f(n) = \left(\frac{1}{(4n+1)^2} + \frac{1}{(4n+2)^2} + \frac{1}{(4n+3)^2}\right)+\left(\frac{1}{(4n+5)^2} + \frac{1}{(4n+6)^2} + \frac{1}{(4n+7)^2}\right)+\cdots+\left(\frac{1}{(8n-3)^2} + \frac{1}{(8n-2)^2} + \frac{1}{(8n-1)^2}\right)$

and

$g(n) = \frac{1}{4n+4}+\frac{1}{4n+8}+\cdots+\frac{1}{8n}$

Taking the absolute value of both sides:

$|s_{8n}-s_{4n}| = |f(n) - g(n)| = |g(n) - f(n)|$

Looking at the terms of $g(n)$, we can see that it has $n$ terms and the smallest term is $\frac{1}{8n}$; so, we can say that $g(n) \geq \frac{n}{8n} = \frac{1}{8}$. So, $\frac{1}{8}$ is a lower bound for $g(n)$. Looking at $f(n)$, we can see that it has $4n - n = 3n$ terms and that its greatest term is $\frac{1}{(4n+1)^2}$. So, we can say that $f(n) \leq \frac{3n}{(4n+1)^2}$. And, since $\frac{3n}{(4n+1)^2}$ is decreasing for $n\geq 1$, it has a maximum at $n = 1$ in this interval; so, substituting $n=1$ into the expression, we get that $f(n) \leq \frac{3}{(4+1)^2} = \frac{3}{25}$. Thus, $\frac{3}{25}$ is an upper bound for $f(n)$. Now, since $\frac{1}{8} > \frac{3}{25}$, the lower bound of $g(n)$ is always greater than the upper bound of $f(n)$ if $n\geq 1$. Therefore, $g(n) > f(n)$ for all $n\geq 1$, and:

$|s_{8n}-s_{4n}| = g(n) - f(n) \geq \frac{1}{8} - \frac{3}{25} = \frac{1}{200}$

The conclusion above is because $g(n)\geq\frac{1}{8}$ and $f(n)\leq \frac{3}{25}$ (so, $-f(n)\geq -\frac{3}{25}$).

So, we can conclude that:

$|s_{8n}-s_{4n}| \geq \frac{1}{200} \text{ whenever } n\geq 1$

But the theorem above requires that, for any $\epsilon>0$, there is a number $N$ such that:

$|s_{8n} - s_{4n}| < \epsilon \text{ whenever } 4n>N$

In particular, if we choose $\epsilon=\frac{1}{200}$, we get a contradiction, because $|s_{8n} - s_{4n}|$ can't get arbitrarily smaller than $\frac{1}{200}$. Thus, the series is divergent.

Is this correct? Is there a different argument?

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    @JulienB., I edited the post to include a second attempt to show that the series is divergent.2012-12-16

1 Answers 1

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A simple proof starts by writing $c_n=a_n-b_n$ with $a_n=\frac1{n^2}$ and $b_n=\left(\frac1{n^2}+\frac1{n}\right)\mathbf 1_{4\mid n}$.

The series $\sum\limits_na_n$ converges absolutely hence $\sum\limits_nc_n$ converges if and only if $\sum\limits_nb_n$ does. Since $\sum\limits_nb_n$ diverges, $\sum\limits_nc_n$ diverges.