I'm taking undergrad real analysis. I'm trying to come up with a general formula for the series \begin{equation} 1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+... \end{equation} and cannot understand how to get the negative term to only show up with powers of $3$. Or is there another way to test the series. any help would be appreciated.
Convergence or divergence of the series $1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...$
2
$\begingroup$
real-analysis
sequences-and-series
convergence
divergent-series
-
0If $F_n$ is the $n$th Fibonacci number with the convention $F_0=0$ and $F_1=1$, then the $n$th term here is $\dfrac{(-1)^{F_n+1}}{n}$. – 2017-02-28
2 Answers
6
The partial sums of this series can be bounded below by partial sums of the series $$1+\frac14+\frac17+\frac1{10}+\cdots$$ which in turn can be bounded below by the partial sums of the series $$\frac13+\frac16+\frac19+\frac1{12}+\cdots=\frac13\left(1+\frac12+\frac13+\frac14+\cdots\right).$$
The latter series is a constant multiple of the harmonic series, hence it diverges.
0
$$ \begin{align} &\quad\, \sum_{n=0}^{\infty}\left[+\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{1}{3n+3}\right] \\[2mm] &= \sum_{n=0}^{\infty}\left[\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{2}{3n+3}+\frac{1}{3n+3}\right] \\[2mm] &= \sum_{n=0}^{\infty}\left[\frac{1}{3n+1}+\frac{1}{3n+2}\color{red}{-}\frac{2}{3n+3}\right]+\sum_{n=0}^{\infty}\left[\frac{1}{3n+3}\right] \\[2mm] &= \log{3}+\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n} \color{red}{\,\,\rightarrow\,\,\infty} \\[2mm] \end{align} $$