"Altered" Alternating Series Diverges or Converges?

Question

Consider the series $$ 1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots $$ which alternates between a block of positives and a block of negatives, with the block sizes, $$ 1, 1, 2, 3, 4, 5, 6, \dots $$ and so on.

Does this series converge or diverge?

Similar questions see here, or here. I suppose your series is also at this site already. — Dietrich Burde, Oct 03 '19 at 20:25
It looks as if the block sizes are $1, 1, 2, 3, 4. 5, \ldots$. Please [edit] the question to clarify. — Ethan Bolker, Oct 03 '19 at 20:26
What precisely is the pattern? We have $1$ positive, $1$ negative, $2$ positives, $3$ negatives, $4$ positives. Will it continue with $5$ negatives, $6$ positives, and so on? — saulspatz, Oct 03 '19 at 20:27
Question edited. Sorry for not being clear enough in the first place. — Niall Taggart, Oct 03 '19 at 20:29
Possible duplicate of Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ — Maximilian Janisch, Oct 03 '19 at 21:01

score 1 · Answer 1 · answered Oct 03 '19 at 20:40

Write $H_n = \sum_{k=1}^{n} \frac{1}{k} $ for the $n$-th harmonic number. It is well-known that $H_n$ has the asymptotic form

$$ H_n = \log n + \gamma + \mathcal{O}\left(\frac{1}{n}\right) \quad \text{as} \quad n\to\infty. $$

Now returning to OP's question, let $s_n$ denote the $n$-th partial sum. Also, write

$$m_k = \frac{k(k+1)}{2} + 1 \qquad \text{and} \qquad K_n = \max\{ k \geq 1 : m_k \leq n \}.$$

Using this, we can write $s_n$ in the the following block form

$$ s_n = 1 + \Bigg( \sum_{k = 1}^{K_n} (-1)^k (H_{m_k} - H_{m_{k-1}}) \Bigg) + (-1)^{K_n+1} (H_n - H_{m_{K_n}} ). $$

Then by using the asymptotic form of $H_n$, we easily find that

$$ H_{m_k} - H_{m_{k-1}} = \frac{2}{k} + \mathcal{O}\left(\frac{1}{k^2}\right), $$

and so,

$$ s_n = 1 + \sum_{k=1}^{K_n} \left( (-1)^{k} \frac{2}{k} + \mathcal{O}\left(\frac{1}{k^2}\right) \right) + \mathcal{O}\left(\frac{1}{K_n}\right). $$

Therefore it follows that $s_n$ converges as $n\to\infty$

score 0 · Answer 2 · answered Oct 04 '19 at 14:17

Let: $$ S=1-\frac{1}{2} +\frac{1}{3} +\frac{1}{4} - \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} + \cdots + \frac{1}{11} + \cdots $$

For a given sequence of additions assume a constant denominator, specifically the smallest in the group. For a series of subtractions pick a common constant denominator, the largest in the group. This guarantees a maximum effect of additions and a minimum effect of subtractions, guaranteeing greater partial sums than the original series.

Doing this we get:

$$S' = 1 -\frac{1}{2}+\frac{2}{3}-\frac{3}{7}+\frac{4}{8}-\frac{5}{17}+\frac{6}{18}...$$

Suppose the numerator is $2k$ for some integer $k$. Then the denominator is $\frac{5k^2-5k+6}{2}$. This term is positive. The term directly to the left is negative with numerator $2k-1$ and denominator $\frac{5k^2-5k+4}{2}$.

So:

$$S'=1 + \sum_{k=1}^\infty \frac{4k}{5k^2-5k+6}-\frac{4k-2}{5k^2-5k+4}$$

This preserves the order of terms.

If the two fractions are added, we get a second degree polynomial in the numerator and a 4th degree polynomial in the denominator for a terms on the order of $O(1/k^2)$.

So $S'$ converges by comparison to $1/k^2$ and $S$ converges by comparison with $S'$.

score 0 · Answer 3 · answered Oct 04 '19 at 14:26

Intuitively:

The blocks of constant sign are made of pieces of the Harmonic series, each with $n$ terms, between $\dfrac{n(n-1)}2$ and $\dfrac{n(n+1)}2$. By the asymptotic formula, we known that these sums are asymptotic to $\log\dfrac{n+1}{n-1}\sim\dfrac2n$. Hence the total sum converges, the same way than the ordinary asymptotic series does.

"Altered" Alternating Series Diverges or Converges?

3 Answers3