# Determine whether $\sum_{n=1}^{+\infty}c_n$ for a given $c_n$ is divergent or convergent

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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You should not use the $\sum_{n=1}^\infty$ symbol since you don't know if the series is convergent (and it's not!) It would be better working with finite sums $\sum_{n=1}^N$ and then check if the sentence is convergent when $N \to \infty$. – Siméon Dec 5 '12 at 12:26
@Julien: I disagree; it's quite usual to use the series symbol without knowing whether the limit exists, just as it is in fact usual to use the limit operator without knowing whether the limit exists, e.g. prototypically in a question like "Does the limit $\lim_{n\to\infty}a_n$ exist?". – joriki Dec 5 '12 at 12:38
The problem here is that it makes the argument fallacious : even if $\sum_{n=0}^\infty a_n$ is convergent, it may be that you cannot split it into packets. For instance think of the alternate harmonic series,is it true that $$\sum_{n=1}^\infty \frac{(-1)^n}{n} = \sum_{n=1}^\infty \frac{1}{2n} - \sum_{n=1}^\infty \frac{1}{2n-1} \quad ?$$ Note that I wouldn't object if the series was non-negative. – Siméon Dec 5 '12 at 12:49
@JulienB.: I think that what makes the expression you wrote to be false is that, if you sum two infinite series term-by-term, and they are both divergent, the result is not necessarily divergent. But the term-by-term sum of a divergent series and a convergent series is necessarily divergent, according to the book I use. So, I think it would not be possible to take the alternate harmonic series and split it into packets such that it becomes a sum of a convergent series and a divergent series, because that would wrongly show that this series is divergent. Or am I thinking wrong? – anonymous Dec 5 '12 at 12:59
@anonymous : You're right, and so is your book. But if you split the series into, let's say 3 series, you can have convergent = convergent + divergent + divergent. All I'm saying is "better be cautious and write only things that really make sense". As long as you know what you do, it's ok. – Siméon Dec 5 '12 at 13:11

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.