# Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-…$

Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ The number of signs increases by one in each "block".

I have an idea. Group the series like this: $1-(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6})-...$

We can show that $1, \frac{1}{2}+\frac{1}{3},\frac{1}{4}+\frac{1}{5}+\frac{1}{6},...$ converges to 0. I'm trying to use Dirichlet's Test. However, I'm not sure wether this sequence is decreasing.

Any idea? Or any other method to establish the convergence?

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Absolute value of sums of blocks is indeed monotone decreasing. Proof is straightforward, just unpleasant to typeset. –  André Nicolas Mar 20 '13 at 17:28
straightforward? Can show $block_{k+1}-block_{k}<0$? –  user67682 Mar 20 '13 at 17:40
Yes, and its's not even close, fairly crude estimates will do it. –  André Nicolas Mar 20 '13 at 17:47

We can see that the general term of this series is $$a_n=(-1)^n\sum_{k=\frac{n(n+1)}{2}+1}^{\frac{n(n+1)}{2}+n+1}\frac{1}{k}$$ and we have $$H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+\frac{1}{2n}+O(\frac{1}{n^2})$$ so \begin{align}|a_n|=H_{\frac{n(n+1)}{2}+n+1}-H_{\frac{n(n+1)}{2}+1}&=\log(1+\frac{2}{n})+\frac{1}{(n+1)(n+2)}-\frac{1}{n(n+1)}+O(\frac{1}{n^2})\\ &=\frac{2}{n}+O(\frac{1}{n^2})\end{align} hence we have $$a_n=\frac{2(-1)^n}{n}+O(\frac{1}{n^2})$$ which allows us to conclude the convergence of the series since it's sum of two convergent series, one by alternating series test and the other by comparaison with Riemann series.

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Replace every $o(1/n^2)$ by $O(1/n^2)$. –  Did Mar 20 '13 at 18:05
@Did thanks for your comment. –  Sami Ben Romdhane Mar 20 '13 at 18:22
And now, why does all this allow us to conclude [that] the convergence of the series [holds]? –  Did Mar 20 '13 at 18:23
@Did The series is sum of two convergent series, one by alternating series test and the other by comparaison with Riemann series. Is not it? –  Sami Ben Romdhane Mar 20 '13 at 18:26
Upvoted, as the only full answer, so far. –  Did Mar 20 '13 at 18:33

If I'm not getting it wrong, your series is

$$1-\frac 1 2 -\frac 13+\frac 14+\frac 15+\frac 16-\frac 17-\frac 18-\frac 19-\frac 1{10}+++++------\dots$$

So you have $1$ plus, $2$ minuses, $3$ pluses, $4$ minuses, and so on.

We can write your series as $a_0+a_1+a_2+a_3+\dots$ where \begin{align} a_0&=(-1)^0 1 \\ a_1&=(-1)^1\sum_{k=2}^{3}\frac 1 k\\ a_2&=(-1)^2\sum_{k=4}^{6}\frac 1 k\\ a_3&=(-1)^3\sum_{k=7}^{10}\frac 1 k\\ \cdots &=\cdots \\ a_n&=(-1)^n\sum_{k=T_n+1}^{T_{n+1}}\frac 1 k\end{align}

Where $T_n=\frac{n(n+1)}{2}$ so it goes $1,3,6,10,\dots$

Now, we know that $$\sum_{k=1}^n \frac 1 k =\log n+\gamma+\frac 1 {2n}+O(n^{-2})$$

Thus

\eqalign{ & \sum\limits_{k = 1}^{{T_n}} {{1 \over k}} = \log n + \log \left( {n + 1} \right) - \log 2 + \gamma + O({n^{ - 2}}) \cr & \sum\limits_{k = 1}^{{T_{n + 1}}} {{1 \over k}} = \log \left( {n + 2} \right) + \log \left( {n + 1} \right) - \log 2 + \gamma + O({n^{ - 2}}) \cr}

Whence, after simplification

$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}} = \log \left( {1 + {2 \over n}} \right) + O(n^{-2})$$

Recall that $$\log(1+x)=x+O(x^2)$$ so

$$\sum\limits_{k = {T_n} + 1}^{{T_{n + 1}}} {{1 \over k}} = \frac{2}{n} +O\left(\frac 1 {n^2}\right)$$

Since $$\sum (-1)^n \frac 1 n$$ and $$\sum n^{-2}$$ converge, so does your series.

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$\sum_{k=1}^{n} \frac{1}{k} = \log n + \gamma + o(1)$. What's this? I'm a year1 maths student. I haven't seen this before... –  user67682 Mar 20 '13 at 17:35
It is a short way of saying that $$\sum_{k=1}^n k^{-1}-\log n\to \gamma$$ when $n\to\infty$; where $\gamma$ is a constant, $0.577...$, known as Euler's constant. Here you can see a proof of that. –  Pedro Tamaroff Mar 20 '13 at 17:38
The last expression proves nothing. By the way, note that $2/n+O(1/n^2)+o(1)$ is simply $o(1)$. –  Did Mar 20 '13 at 18:02
@Did What improvement do you suggest? –  Pedro Tamaroff Mar 20 '13 at 18:09
PeterT: The proof is correct now. +1. –  Did Mar 20 '13 at 19:40

You can use the alternating series test if you can prove that the sum of a block goes to zero. Block $n$ starts at $\dfrac 1{\dfrac {n(n-1)}2+1}$ and ends at $\frac 1{\frac {n(n+1)}2}$ and has $n$ terms. The sum is then less than $\dfrac n{\dfrac {n(n-1)}2}=\dfrac 2{n-1}$ which goes to zero

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You have also to prove the block is decreasing. –  Sami Ben Romdhane Mar 20 '13 at 17:27
I have proved the sum of a block converges to 0. But alternating series test also requires the sum of block is a decreasing sequence. I failed to prove the sequence is decreasing... –  user67682 Mar 20 '13 at 17:27
The same argument shows the sum of a block is greater than $\frac 2{n+1}$. A little finer work will show it is $\frac 2n+O(\frac 1{n^2})$, which is nicely decreasing. –  Ross Millikan Mar 20 '13 at 17:33
@RossMillikan See my answer. –  Pedro Tamaroff Mar 20 '13 at 17:57
@RossMillikan: $\frac2n+O\left(\frac1{n^2}\right)$ does not ensure monotonicity. For example $\frac2{n-(-1)^n}$ is $$\left\{1,2,\frac12,\frac23,\frac13,\frac25,\dots\right\}$$ –  robjohn Mar 20 '13 at 18:19

$$\underbrace{\vphantom{\frac11}+1}_{\text{length }1} \underbrace{-\frac12-\frac13}_{\text{length }2} \underbrace{+\frac14+\frac15+\frac16}_{\text{length }3} \underbrace{-\frac17-\frac18-\frac19-\frac1{10}}_{\text{length }4} \underbrace{+\frac1{11}+\frac1{12}+\frac1{13}+\frac1{14}+\frac1{15}}_{\text{length }5}-\ldots$$ The absolute values of the terms of the same-sign block of length $n$ are from $$\dfrac1{n(n-1)/2+1}\quad\text{to}\quad\dfrac1{n(n+1)/2}$$ and the sum of the block must satisfy $$\frac2{n+1}=\frac{n}{n(n+1)/2}\le(-1)^{n-1}\text{sum}\le\frac{n}{n(n-1)/2+1}\lt\frac2{n-1}$$ which tends to $0$.

The absolute value of the sum of the same-sign block of length $n$ and the same-sign block of length $n+1$ is at most $$\frac2{n-1}-\frac2{n+2}=\frac6{(n+2)(n-1)}$$ Thus, the sum of pairs of blocks converge absolutely and the blocks converge to $0$. Thus, the full series converges.

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Very nice... +1 –  user1551 Mar 20 '13 at 19:07

To elaborate on Ross Millikan's answer and André Nicolas's comment, denote the $n$-th block sum (ignoring the sign) by $B_n$. Then \begin{align} B_n &= \sum_{i=1}^n \frac1{k_n+n},\\ k_n &= \frac{n(n-1)}{2},\\ k_{n+1} &= k_n+n = \frac{n(n+1)}{2} \le n^2.\tag{1} \end{align} Therefore \begin{align} B_n - B_{n+1} &= \sum_{i=1}^n \frac1{k_n+i} - \sum_{i=1}^{n+1} \frac1{k_{n+1}+i}\\ &= \sum_{i=1}^n \frac1{k_n+i} - \sum_{i=1}^n \frac1{k_{n+1}+i} - \frac1{k_n+2n+1}\\ &= \sum_{i=1}^n \frac{n}{(k_n+i)(k_{n+1}+i)} - \frac1{k_n+2n+1}\ \text{ by } (1)\\ &\ge \sum_{i=1}^n \frac{n}{(k_n+n)(k_n+2n)} - \frac1{k_n+2n+1}\\ &= \frac{n^2}{(k_n+n)(k_n+2n)} - \frac1{k_n+2n+1}\\ &\ge \frac{1}{k_n+2n} - \frac1{k_n+2n+1}\ \text{ by } (1)\\ &> 0. \end{align} Hence the block sum is indeed monotonic decreasing and the alternating series test applies.

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I figured I'd take a different approach to it, by providing upper and lower bounds.

For an upper bound, we'll assume that all values in the positive sequences are the largest of them, and all values in the negative sequences are the smallest. That is, for instance...

$\frac{1}{4} + \frac{1}{5} + \frac{1}{6} < \frac{3}{4}$

And

$\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}>\frac{4}{10} = \frac{2}{5}$

For the positive terms, we have that the sums are bounded above by $\frac{2n+1}{n(2n+1)+1} < \frac{1}{n}$, while the negative terms' sums are bounded below by $\frac{2}{2n+1} > \frac{1}{n+1}$. As such, we can say easily that the series cannot be larger than

$$1+\sum_{n=1}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right) = 2$$ Similarly, the positive sums are bounded below by $\frac{1}{n+1}$ and the negative sums are bounded above by $\frac{2n}{n(2n-1)+1} < \frac{1}{n-1}$. And so, the series cannot be smaller than

$$\frac{1}{2}+\sum_{n=2}^\infty \left(\frac{1}{n+1} - \frac{1}{n-1}\right) = -1$$

As such, we know that the series must be less than 2 and greater than -1, so we know that the series does not diverge, and thus we only need to show that it does not fail to converge due to oscillation. As both the positive and negative sum terms behave as $1/n$ in the limit as $n \to \infty$ (thereby preventing cyclic behaviour), we know that the series must converge.

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