Convergent/divergent series Is the following series divergent/convergent? 
$$S=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}-...$$
I think it is divergent since
$$
\begin{align}
S&>1-\frac{1}{2}-\frac{1}{2}+4\cdot\frac{1}{6}-\frac{4}{7}+\frac{5}{15}-\frac{6}{16}+...\\
&=1/3-4/7+1/3-6/16+...=\sum_{n=1}^\infty 1/3-\alpha_n
\end{align}
$$
where $\alpha_n=4/7, 6/16, 8/22$ which tends to 0, so the series on the right hand side is divergent. Is this the right answer? Thanks 
 A: After placing brackets around the terms with the same sign the series becomes
$$1+\sum_{n=2}^\infty(-1)^{n+1}(H_{n(n+1)/2}-H_{n(n-1)/2}),$$
where $H_n=1+\frac12+\frac13+\ldots+\frac1n$ is the $n$-th harmonic number. Now using well-known formula $$H_n=\log n+\gamma+O\left(\frac{1}{n}\right),$$
we get
$$a_n=H_{n(n+1)/2}-H_{n(n-1)/2}=\log\frac{n+1}{n-1}+O\left(\frac{1}{n^2}\right)=$$
$$=\log\left(1+\frac{2}{n-1}\right)+O\left(\frac{1}{n^2}\right)=\frac{2}{n}+O\left(\frac{1}{n^2}\right).$$
Therefore the main term of the series with braces is $$2\frac{(-1)^{n+1}}{n}+O\left(\frac{1}{n^2}\right),$$
which means that the series converges (by Leibniz's and comparison tests), and so does the original one.
A: You are taking groups of terms (with the same sign) $n$ terms at a time, for $n=1,2,\ldots$. This means that the absolute value of sum of the $n$th group of terms is basically $H({{n+1} \choose 2}) - H({n \choose 2})$, where $H(k)$ is the $k$th partial harmonic sum, and due to the very strong convergence of $H(k)$ to $\log k$, this means that your groups of $n$ terms are very closely proportional (in absolute value) to $|\log (n+1) - \log (n - 1)|$, with constant of proportionality $1$, and thus you get an "approximate" alternating series, when you consider ratios of aggregated $n$ terms to $|\log (n+1) - \log (n - 1)|$. I've left out some details, but the main point is that $\lim_{k \to \infty} \log k - H(k)$ is a constant, which should be most of what you need to turn this into a proof that the series is convergent, basically because alternating series are convergent when their terms are strictly decreasing to zero.
A: To provide further insight to Santu's answer, some details.
$S$ is not per se an alternating series. But with some work, we can cast it into one. Note that, modulo the sign, each block $b_n$ of $n$ term with the same sign can be expressed as
$$
b_n = \sum_{\frac{n(n+1)}{2}+1}^{\frac{(n+1)(n+2)}{2}}\frac{1}{i}
$$
or equivalently
$$
b_n = \sum_1^{\frac{(n+1)(n+2)}{2}}\frac{1}{i}-\sum_1^{\frac{n(n+1)}{2}}\frac{1}{i}
$$
as you can check by hand. Then your series can be written as $S=\sum_{n=0}^\infty (-1)^n b_n$. In order to show that it is conditionally convergent you must show three facts


*

*that is alternating

*that $b_n\to 0$ as $n\to \infty$

*that $b_{n+1}\leq b_n$, at least for $n$ large enough.


The last point is often forgotten and it is crucial (see the final note). So, clearly the series is alternating. For $b_n$ we have the bound obtained by replacing each term in the sum with the largest one
$$
b_n \leq \frac{2n-2}{n(n+1)+2}
$$
By the squeeze theorem, $b_n\to 0$ as $n\to\infty$. Finally, is $b_n$ decreasing? I did some calculation and I think the answer is yes. To check it, consider $b_n - b_{n+1}$, and use the bound (obtained with comparison with $\int_1^n \frac{1}{x}dx$)
$$
\ln(n+1) < \sum_1^n \frac{1}{i} < \ln(n) + 1
$$
The math is not hard, but it could get long here. You can do it yourself, and try to bound $b_n-b_{n+1}$ from below (careful with signs!).
Final note: as for the counter example of what goes wrong if you miss the 3rd requirement, consider the series $\sum (-1)^n a_n$, with $a_n$ given by
$$
a_n =
\begin{cases}
\frac{1}{n}\quad \mbox{ for $n$ odd}\\
0 \quad \mbox{ for $n$ even}.
\end{cases}
$$
You can see that such series diverges.
A: We should be careful here. It's tempting to group the terms into natural subsums and argue the grouped series converges and leave it at that. But the convergence of a grouped series converges does not imply the original series converges in general. The obvious example being $1 + (-1) + 1 + (-1) +\cdots.$  Here's an example with the terms $\to 0$:
$$1-1 +1/2+1/2 -1/2 - 1/2 + 1/3+1/3 +1/3 - 1/3 -1/3-1/3 + \cdots.$$
One sufficient condition for the convergence of the grouped series to imply convergence of the orignal series is the following: Suppose the original series is $\sum_n a_n$ and we know the grouped series
$$(a_1 + \cdots + a_{n_1}) + (a_{n_1+1}+ \cdots +a_{n_2 }) + \cdots $$
converges. If
$$\lim_{k\to\infty} (|a_{n_k+1}|+ \cdots +|a_{n_{k+1}}|) = 0,$$
then the original series converges. I'll leave the proof to the reader as it's not too bad. (Briefly, the condition shows the partial sums of the original series differ from those of the grouped series by an amount $\to 0,$ giving the desired convergence.)
For the problem at hand, this is fairly easy to check.

Here's my own proof for this specific problem, where I don't appeal to the $\log $ function at all. For the $k$th grouping of positive and negative terms, let $M=(2k-2)(2k-1)/2, N = (2k-1)(2k)/2.$ Then the grouping is
$$\tag 1 \frac{1}{M+1} + \cdots + \frac{1}{M+2k-1} - \frac{1}{N+1}-\cdots - \frac{1}{N+2k}$$ $$ = \sum_{j=1}^{2k-1} \frac{N-M}{(M+j)(N+j)} - \frac{1}{N+2k}.$$
Note that $N-M = 2k-1.$ In absolute value then, making absolutely brazen estimates, $(1)$ is $\le (2k-1)^2/(M+1)^2 + 1/N.$ This is on the order of $1/k^2$ as $k\to \infty.$ This implies the grouped series converges absolutely, hence converges. How about the original series? It converges by the first part of this answer, because if we slap absolute values on each term in $(1),$ the sum is is no more than $(4k-1)/M+1 \to 0$ as $k\to 0.$ This finishes the proof.
A: S is convergent ( Using Leibnitz's Test for convergence of Alternating series)
