Investigate convergence of the following series Investigate the corvergence of the following series?
$1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\frac{1}{13}+\frac{1}{15}-\frac{1}{8}+ \ldots$
$1+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}-\frac{1}{15}+ \ldots$
I need a hit about which cirterion to use in each of those cases and i do not know how to find a general term of those twos sequence?
 A: $$\begin{eqnarray*}S_1&=&\sum_{n\geq 0}\left(\frac{1}{4n+1}+\frac{1}{4n+3}-\frac{2}{4n+4}\right)=\sum_{n\geq 0}\int_{0}^{1}\left(x^{4n}+x^{4n+2}-2 x^{4n+3}\right)\,dx\\&=&\int_{0}^{1}\frac{1+x^2-2x^3}{1-x^4}\,dx=\int_{0}^{1}\frac{dx}{1+x}+\int_{0}^{1}\frac{x\,dx}{1+x^2}=\log 2+\frac{1}{2}\log 2\\&=&\color{red}{\frac{3}{2}\log 2}.\end{eqnarray*}$$
In general, the Dirichlet's criterion gives that, if $\{a_n\}$ is a sequence with bounded partial sums and $\{b_n\}$ is a decreasing sequence converging to zero, then the series $\sum_{n\geq 1}a_n b_n$ converges (proof: partial summation). In the first case, if we take $\{a_n\}=\{1,1,0,-2,1,1,0,-2,\ldots\}$ and $b_n=\frac{1}{n}$ we have that $S_1$ is convergent.
In the second case, if we take $\{a_n\}=\{1,1,-1,-1,0,1,1,-1,-1,0\}$ and $b_n=\frac{1}{n}$ we have that
$$ T_2 = \sum_{n\geq 0}\left(\frac{1}{5n+1}+\frac{1}{5n+2}-\frac{1}{5n+3}-\frac{1}{5n+4}\right)$$
converges to some constant (for instance, $\frac{\pi}{5}\sqrt{2+\frac{2}{\sqrt{5}}}=1.068959$ by the previous method), hence
$$ S_2 = \sum_{n\geq 0}\left(\frac{1}{5n+1}+\frac{1}{5n+2}-\frac{1}{5n+3}-\frac{1}{5n+4}+\frac{1}{5n+5}\right)$$
diverges.
A: For the second, compute
$$\frac1{5n+1}+\frac1{5n+2}-\frac1{5n+3}-\frac1{5n+4}-\frac1{5n+5}=\frac{-625n^4+\cdots}{3125n^5+\cdots}$$
and compare with harmonic series.
A: For the first: rearrange the terms to get $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}+\frac{1}{11}+\frac{1}{13}+\frac{1}{15}+ \ldots$$ Note that the first part of this (until $+\frac{1}{9}$) is exactly the alternating harmonic series, which is known to converge. However, there remain missing three even terms ($-\frac{1}{10}, -\frac{1}{12},-\frac{1}{14}$). In fact, there will always be even terms missing since the series you gave has two odd terms for every even term. Furthermore, as the series goes on the number of stray odd terms increases to infinity. So we know the first "half" of the series converges, since it is the alternating harmonic series, the question then becomes: does the harmonic series of only odd numbers converge or diverge?
A: For the first one, you have consecutive triples
$$\frac1{4n-1}-\frac1{2n}+\frac1{4n+1}=\frac{1}{2n(4n-1)(4n+1)}\sim\frac1{32n^3}$$
The sum is thus convergent.
For the record, the denominators of the terms in the sum are, up to sign, this sequence in OEIS.
