Proof verification $\sum_{n=1}^{\infty} (-1)^n \frac{1}{n}$ converges My proof:
The convergence of $\sum_{n=1}^{\infty} (-1)^n \frac{1}{n}$ is equivalent to $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}$ converging.
\begin{equation}
\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n} = \frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \cdots = \frac{1}{1\cdot 2}+\frac{1}{3\cdot 4} + \cdots =\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n)}< \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}< \sum_{n=1}^{\infty} \frac{1}{n^2} 
\end{equation}
It is well know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, and in the third term of the equation we see, that it is strictly increasing, hence it must also converge. QED.
I would like to know, whether or not this proof is correct, and how I could make it more professional? Maybe rigorous is the word I'm looking for
 A: No, it is not correct. You are assuming that a series$$a_1+a_2+a_3+a_4+a_5+a_6+\cdots$$converges if and only if the series$$(a_1+a_2)+(a_3+a_4)+(a_5+a_6)+\cdots$$converges. This is not true. For instance, the series$$1-1+1-1+1-1+\cdots$$diverges, whereas the series$$(1-1)+(1-1)+(1-1)+\cdots$$converges. I suggest that you apply the alternating series test here.
A: As said above, you can not regroup the terms of the series 2 by 2. Other possibilities for showing that this series converges are :

*

*Using the alternating series test

*Try to find a more elementary proof

As the one you proposed is not good (because the series converges conditionally and not absolutely), I propose you this proof that does not use the criteria :
\begin{align}
\sum_{n=1}^{N}\frac{(-1)^{n+1} }{n} &=  -\sum_{n=1}^{N} \frac{(-1)^{n} }{n} \\
&=-\sum_{n=1}^{N}(-1)^{n}\int_0^1 x^{n-1} dx  \\
&= \int_0^1 \sum_{n=1}^{N} (-x)^{n-1} dx \\
&= \int_0^1 \frac{1-(-x)^{N}}{1-(-x)} dx \\
&= \int_0^1 \frac{1}{1+x} dx - \int_0^1 \frac{(-x)^{N}}{1+x}dx \\
&= \ln(2) - \int_0^1 \frac{(-x)^{N}}{1+x}dx \\
\end{align}
Then we can show that the second integral converges towards $0$ when $N$ grows to infinity :
$$ |\int_0^1 \frac{(-x)^{N}}{1+x}dx| \leq \int_0^1 x^{N}dx = \frac{1}{N+1} \to 0$$
A: As pointed out, convergence of $(a_1+a_2)+(a_3+a_4)+\dots$ does not imply convergence of $\sum a_n$. Except it's easy to show that here it does:


Prop. If $(a_1+a_2)+(a_3+a_4)+\dots$ converges and $a_n\to0$ then $\sum a_n$ converges.


Partial Proof: Say $s_n=a_1+\dots a_n$ and $\sigma_k=(a_1+a_2)+\dots+(a_{2k-1}+a_{2k})$. We need to show that if $\sigma_k\to s$ and $a_n\to0$ then $s_n\to s$.
I leave the actual proof of that to you; it begins "Let $\epsilon>0$".
Note that if $n=2k$ then $s_n=\sigma_k$, while if $n=2k+1$ then $s_n=\sigma_k+a_{2n+1}$.
