Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with the same sing. 
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$
I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.
I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$
Cant find a way to proceed.
Thanks for helping.
 A: So we have to deal with
$$ S=\sum_{k\geq 1}(-1)^{k+1} \sum_{n=\binom{k}{2}+1}^{\binom{k+1}{2}}\frac{1}{n} = \sum_{k\geq 2} (-1)^k A_k $$
and to prove convergence it is enough to show that $\{A_k\}_{k\geq 1}$ is decreasing (from some point on) and convergent to zero. The last claim is straightfoward to prove, since $A_k\geq 0$ but 
$$A_k\leq \frac{k}{\binom{k}{2}+1}\sim\frac{2}{k}. $$
The first claim can be easily proved by induction or a convexity argument:
$$ A_k = H_{\binom{k+1}{2}}-H_{\binom{k}{2}}, $$
hence $\{A_k\}$ is decreasing iff $k\mapsto H_{\binom{k}{2}}$ is a midpoint-concave function. Since:
$$\begin{eqnarray*} H_n=\sum_{k=1}^{n}\frac{1}{k} &=& \sum_{k=1}^{n}\left(\frac{1}{k}-\log\frac{k+1}{k}\right)+\sum_{k=1}^{n}\log\frac{k+1}{k}\\&=&\gamma+\log(n+1)-\sum_{k>n}\left(\frac{1}{k}-\log\frac{k+1}{k}\right)\end{eqnarray*}$$
and $\left(\frac{1}{k}-\log\frac{k+1}{k}\right)$ is bounded between $\frac{1}{2k(k+1)}$ and $\frac{1}{2k^2}$, we have:
$$ H_{\binom{k+1}{2}}-H_{\binom{k}{2}} = \log\frac{k^2+k+2}{k^2-k+2}+O\left(\frac{1}{k^4}\right)$$
so $x\mapsto H_{\binom{k}{2}}$ is a midpoint-concave function from some point on.
We may also notice that the value of our series depends on the integral:
$$ I =\int_{0}^{1}\frac{1+2\sum_{k\geq 1}(-1)^k x^{\binom{k}{2}}}{x-1}\,dx.
 $$
A: Michael's grouping answer works in general.  Note that the group with $n$ terms is of the form
$${1\over k+1}+{1\over k+2}+\cdots+{1\over k+n}$$
so the next group is
$${1\over k+n+1}+{1\over k+n+2}+\cdots{1\over k+2n}+{1\over k+2n+1}$$
It's easy to see that
$${1\over k+j}-{1\over k+n+j}={n\over(k+j)(k+n+j)}\gt{n\over(k+n)(k+2n+1)}$$
for each $1\le j\le n$, and so to show that the group with $n$ terms is bigger than the next group, it suffices to show
$${n^2\over(k+n)(k+2n+1)}\ge{1\over k+2n+1}$$
which is to say,
$$n^2\ge k+n$$
But this is easy:  The number that ends the group with $n$ terms, $k+n$, is the triangular number $n(n+1)\over2$.  And $n^2\ge{n(n+1)\over2}$ for all $n\ge1$.
A: Compare the absolute value of each homogeneously signed sum with the sum containing an equal number of copies of the first term.  For example, the all-negative terms from $-1/7$ to $-1/10$ are compared with four copies $-1/7$.  The absolute values of the sums then satisfy
$(1/7+1/8+1/8+1/9+1/10)<4/7$
In general the cluster sums have absolute values less than 
$n/({n(n-1)}/2+1)$
which goes to zero as $n$ increases without bound, so the series passes the test for convergence.
A: Compare each group, term by term, with the next group.
$$\frac14-\frac17+\frac15-\frac18+\frac16-\frac19=\frac3{4\cdot7}+\frac3{5\cdot8}+\frac3{6\cdot9}>\frac{3^2}{6\cdot9}>\frac19>\frac1{10}$$
Each group is greater than the next group, so the alternating series test applies.
A: This answer uses similar ideas to those in Jack D'Aurizio's solution:
First we let $\displaystyle c_n=\frac{n^2-n+2}{2}$ and let $\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}$  for each n,
and we consider the related series $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$ obtained by grouping terms with the same sign,
so $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n=1-\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)+\cdots$
1) $\;\;\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}\le\sum_{k=0}^{n-1}\frac{1}{c_n}=\frac{n}{c_n}=\frac{2n}{n^2-n+2}\to0,\;\;$ so $a_n\to 0$.
2) $\;\;\displaystyle a_n=\sum_{k=0}^{n-1}\frac{1}{c_n+k}>\int_{c_n}^{c_n+n}\frac{1}{x}\;dx=\ln\left(1+\frac{2n}{n^2-n+2}\right)$ for each n, and
$\;\;\;\;\displaystyle a_{n+1}=\sum_{k=0}^{n}\frac{1}{c_{n+1}+k}<\int_{c_{n+1}-1}^{c_{n+1}+n}\frac{1}{x}\;dx=\ln\left(1+\frac{2}{n}\right)$ for each n.
$\hspace{.2 in}$Since $\displaystyle\frac{2n}{n^2-n+2}\ge\frac{2}{n}$ for $n\ge2$, $\;\;a_n > a_{n+1}$ for $n\ge 2$.
Therefore $\displaystyle \sum_{n=1}^{\infty}(-1)^{n+1}a_n$ converges by the Alternating Series Test.

Let $(T_n)$ be the sequence of partial sums of this series, and let $T_n\to T$.
If $(S_k)$ is the sequence of partial sums for the original series,
then for any value of $k$ let $n$ be the largest integer satisfying $n^2-n+2\le 2k$.
Then $\big|S_k-T\big|\le\max\{|T_n-T|, |T_{n+1}-T|\}$, so it follows that $S_k\to T$.
A: 1.in case of non absolute convergence, grouping is not a good idea.
2.Sum $S, \quad \frac {1}{2}<S<1$, using Bolzano-Weisrass theorem convergence follows
