I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott. I am having some trouble determining the convergence or divergence of the below infinite series. I have some initial thoughts which I have put down. Any inputs would be really helpful.
How do I build more skill at finding out if a given infinite series is convergent/divergent?
Exercise 2.7.2
(c) $1 – \frac{3}{4} + \frac{4}{6} – \frac{5}{8} + \frac{6}{10} – \frac{7}{12} + \ldots$
(d) $1 + \frac{1}{2} – \frac{1}{3} + \frac{1}{4} + \frac{1}{5} – \frac{1}{6} + \frac{1}{7} + \frac{1}{8} – \frac{1}{9} + \ldots$
(e) $1 – \frac{1}{2^2} + \frac{1}{3} – \frac{1}{4^2} + \frac{1}{5} – \frac{1}{6^2} + \frac{1}{7} – \frac{1}{8^2} + \ldots$
Proof.
(c) Let $(s_n)$ be the sequence of partial sums of the infinite series $\sum_{n=1}^{\infty}a_n$.
The general expression for the infinite series is,
\begin{align*}
1 + \sum_{n=1}^{\infty}(-1)^{n}\cdot\frac{n+2}{2n}
\end{align*}
I cannot apply the alternate series test, because
\begin{align*}
\lim_{n\to\infty}a_n = \lim_{n\to\infty}\frac{n+2}{2n} = \frac{1}{2} \ne 0
\end{align*}
I can't think of a known convergent or divergent series to compare with, in order to use the Alternate Series test.
(d) One observation is, this is not a rearrangement of the alternating harmonic series.
Best Answer
(c) Your series is $\sum_{n=1}^\infty(-1)^{n+1}\frac{n+1}{2n}$ and, since you don't have $\lim_{n\to\infty}(-1)^{n+1}\frac{n+1}{2n}=0$, the series diverges.
(d) The sum of the first $3$ terms is greater than $1$. The sum of the first $6$ terms is greater than $1+\frac14$. The sum of the first $9$ terms is greater than $1+\frac14+\frac17$. So, your series diverges.
(e) The sum of the positive terms diverges, whereas the sum of the negative terms converges, and therefore this series diverges.