Sum of "alternating-like" series I tried to prove that the following series converges.
$$1 - \frac{1}{2} - \frac{1}{3} + \frac{1}{4}+ \frac{1}{5} - \frac{1}{6} - \frac{1}{7} + \frac{1}{8} +\frac{1}{9} - ...$$
I found a expression for this sum in sigma form by summing pair of numbers with the same sign:
$$1 + \sum_{n=1}^{\infty} \frac{8n +1}{16n^{2}+4n} - \frac{8n -3}{16n^{2} - 12n +2} = 1 + \sum_{n=1}^{\infty} \frac{(8n +1)(16n^{2}+4n) - (8n -3)(16n^{2}+4n)}{(16n^{2}+4n)  (16n^{2} - 12n +2)}$$
But I'm getting a very long expression that can't find a way to prove a convergence of it , I feel that there is more elegant way to prove convergences of this series.
I will thank for your help.
 A: When you have a series with positive and negative numbers, no harm is done if you put parenthesis around groups of terms with the same sign. So, your series converges if and only if the series$$1-\left(\frac12+\frac13\right)+\left(\frac14+\frac15\right)-\cdots$$converges. Which it does, by the alternating series test.
A: This follows by alternating series test..
Suppose that we have a series $\sum a_n$ and either $a_n = (-1)^n b_n$ or $a_n = (-1)^{n + 1} b_n$, where $b_n \geq 0$ for all n.
Assume that the following conditions hold:

*

*$\lim\limits_{n \rightarrow \infty} \ b_n = 0$


*$\{ b_n \}$ is a decreasing sequence, i.e.
$ b_1 \geq b_2 \geq b_3 \cdots \geq b_n \geq \cdots $
Then the alternating series $\sum a_n$ is convergent.
Note that the series
$
1 - \left( {1 \over 2} + {1 \over 3} \right) +
\left( {1 \over 4} + {1 \over 5} \right) - \cdots
$
satisfies the two conditions of the alternating series test.
Hence, it follows that it is convergent. $\blacksquare$
A: Your approach is indeed correct.
So the series in question is equal to
$$1+ \sum_{n=1}^{\infty} \frac{8n +1}{16n^{2}+4n} - \frac{8n -3}{16n^{2} - 12n +2} = 1+ \sum_{n=1}^{\infty} \frac{-32 n^2+8 n+1}{128 n^4-64 n^3-8 n^2+4 n}. 
$$
Note that in the last expression for the series the terms have a definite sign (in particular they are all negative). In this form the series converges because for large $n$ the terms behave as
$
a_n =O(1/n^2)$ (the numerator grows as $n^2$ while the denominator grows as $n^4$) and so it converges by the comparison test if you want.
A: $$S_p=1 + \sum_{n=1}^{p} \frac{8n +1}{16n^{2}+4n} - \sum_{n=1}^{p}\frac{8n -3}{16n^{2} - 12n +2} $$
Notice that
$$ \frac{8n +1}{16n^{2}+4n}=\frac{1}{4 n+1}+\frac{1}{4 n}$$
$$\frac{8n -3}{16n^{2} - 12n +2}=\frac{1}{4 n-1}+\frac{1}{4n-2}$$ Computing the four summations
$$\sum_{n=1}^{p}\frac{1}{4 n+1}=\frac{1}{4} \left(\psi \left(p+\frac{5}{4}\right)-\psi \left(\frac{5}{4}\right)\right)$$
$$\sum_{n=1}^{p}\frac{1}{4 n}=\frac{1}{4} H_p$$
$$\sum_{n=1}^{p}\frac{1}{4 n-1}=\frac{1}{4} \left(\psi \left(p+\frac{3}{4}\right)-\psi \left(\frac{3}{4}\right)\right)$$
$$\sum_{n=1}^{p}\frac{1}{4n-2}=\frac{1}{4} \left(\psi\left(p+\frac{1}{2}\right)-\psi\left(\frac{1}{2}\right)\right)$$ Summing and using harmonic numbers in place of digamma's
$$S_p=\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{1}{4} \left(-H_{p-\frac{1}{2}}-H_{p-\frac{1}{4}}+H_p+H_{p+\frac{1}{4}}\right)$$ Using the asymptotics
$$S_p=\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{1}{4 p}-\frac{3}{32 p^2}+O\left(\frac{1}{p^3}\right)$$gives the limit, shows how the limit is approached and provides a shortcut method for the evaluation of the partial sums.
Using $p=10$
$$S_{10}=\frac{101405260644075743}{219060189739591200}=0.462910\cdots$$ while the truncated expansion gives
$$S_{10}\sim\frac{\pi }{4}-\frac{\log (2)}{2}+\frac{77}{3200}=0.462887\cdots$$ to which corresponds a relative error of $0.005$%.
A: Grouping
Since the terms tend to zero, we can group them in pairs:
$$
\begin{align}
1-\frac12-\frac13+\frac14+\frac15-\frac16-\frac17+\frac18+\dots
&=\sum_{k=1}^\infty(-1)^{k-1}\left(\frac1{2k-1}-\frac1{2k}\right)\tag{1a}\\
&=\sum_{k=1}^\infty(-1)^{k-1}\frac1{(2k-1)2k}\tag{1b}
\end{align}
$$
The series in $\text{(1b)}$ converges by the Alternating Series Test. Furthermore, the series in $\text{(1b)}$ converges absolutely due to the convergence of
$$
\begin{align}
\sum_{k=1}^\infty\frac1{(2k-1)2k}
&=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)\tag{2a}\\
&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\tag{2b}
\end{align}
$$
which also converges by the Alternating Series Test.
Similarly, we can group terms in quadruples:
$$
\begin{align}
&\sum_{k=1}^\infty\left(\frac1{4k-3}-\frac1{4k-2}-\frac1{4k-1}+\frac1{4k}\right)\tag{3a}\\
&=\sum_{k=1}^\infty\left(\frac1{(4k-3)(4k-2)}-\frac1{(4k-1)4k}\right)\tag{3b}\\
&\le\sum_{k=1}^\infty\left(\frac1{(4k-3)(4k-2)}-\frac1{(4k+1)(4k+2)}\right)\tag{3c}\\[3pt]
&=\frac12\tag{3d}
\end{align}
$$
Explanation:
$\text{(3a)}$: sum the series in groups of four
$\text{(3b)}$: collect the first two and last two terms together
$\phantom{\text{(3b):}}$ in this form, we can see that the sum of the group of four terms is positive
$\text{(3c)}$: increase the difference by reducing the subtrahend
$\text{(3d)}$: the series telescopes
When a sum of positive terms is bounded above, it converges.

Evaluating
If both
$$
\sum_{k=1}^\infty a_k\quad\text{and}\quad\sum_{k=1}^\infty b_k\tag4
$$
converge, then
$$
\sum_{k=1}^\infty(a_k+b_k)=\sum_{k=1}^\infty a_k+\sum_{k=1}^\infty b_k\tag5
$$
We can apply $(5)$ to the series in $\text{(1a)}$, which breaks up the original series into even and odd terms:
$$
\begin{align}
\sum_{k=1}^\infty(-1)^{k-1}\left(\frac1{2k-1}-\frac1{2k}\right)
&=\overbrace{\sum_{k=1}^\infty\frac{(-1)^{k-1}}{2k-1}}^\text{odd terms}+\overbrace{\sum_{k=1}^\infty\frac{(-1)^k}{2k}}^\text{even terms}\tag{6a}\\
&=\frac\pi4-\frac12\log(2)\tag{6b}
\end{align}
$$
Both of the series on the right hand side in $\text{(6a)}$ converge by the alternating series test. The series of odd terms is the Leibniz Formula for $\pi$, whose sum is $\frac\pi4$, and the series of even terms is $-\frac12$ times the Alternating Harmonic Series, whose sum is $\log(2)$.
We have $\frac\pi4-\frac12\log(2)\approx0.438824573117475655$, which is less than $\frac12$, in agreement with $(3)$.
