Limit and Convergence of an Alternating Series

Question

Without applying the theorem using the $lim b_n = 0$, how can we prove that this Alternating Series diverges?

$$\sum_{n=1}^\infty (-1)^{n-1}\frac{n+1}{n}$$

The ratio and root tests are inconclusive $L = 1$

Can I use the comparison test with $\sum_{n=1}^\infty \frac{1}{n}$ which diverges? But I am worried about the negative terms in the series.

Finally, computationally we have $$\lim_{n\to\infty}(-1)^{n-1}\frac{n+1}{n} = (1 + \frac{1}{n}) e^{-i π + i n π} $$

Answer 1

Because if $\sum a_n$ converges, $lim_na_n=0$

Answer 2

Without using $\lim b_n = 0$, we can argue as follows:

Let $S_n$ denote the $n$-th partial sum of the sequence. Then $$S_n = (1-1+1-1+...+(-1)^{n-1}) + (1-\frac12+\frac13-\frac14+...+(-1)^{n-1}\frac1n) = G_n + K_n$$

where $G_n$ is $0$ for $n$ even and $1$ if $n$ is odd, and $K_n = 1-\frac12+\frac13-\frac14+...+(-1)^{n-1}\frac1n$. Let $m$ be an arbitrary positive integer. Then

$$K_{2m} = (1-\frac12)+(\frac13-\frac14)+...+(\frac{1}{2m-1}-\frac{1}{2m}) = \frac12 + \frac{1}{12} + \frac{1}{30} +...+\frac{1}{(2m-1)(2m)}.$$

We may use induction on $m$ to see that $K_{2m} < 1-\frac{1}{3m}$, so that $K_{2m} < 1.$ On the other hand, $K_{2m} > \frac12$. By considering $G_{2m} = 0$ along with the just obtained lower and upper bounds for $K_{2m}$, we see that $1 > S_{2m} > \frac12$ so that $1\geq \lim S_{2m} \geq \frac12.$ But

$$K_{2m+1} = 1+(-\frac12 + \frac13) + (-\frac14+\frac15)+(-\frac16+\frac17)+...+(-\frac{1}{2m}+\frac{1}{2m+1}) = 1-\frac16 - \frac{1}{20}-\frac{1}{42}-...-\frac{1}{(2m)(2m+1)}$$

and we may use induction to show that $K_{2m+1} > \frac12 + \frac{1}{4m}$, and hence $K_{2m+1} \geq \frac12.$ By considering $G_{2m+1} = 1$ along with the just found lower bound for $K_{2m+1}$, we see that $S_{2m+1} > \frac32$, and hence $\lim S_{2m+1} \geq \frac32$. Thus, the even limit of $S_n$ and the odd limit of $S_n$ cannot be the same, and thus the limit doesn't exist.

The way I found out the crucial steps $K_{2m} < 1-\frac{1}{3m}$ and $K_{2m+1} > \frac12 + \frac{1}{4m}$ is that I first wrote out first few terms of the corresppnding subsequences, and noticed that $K_{2m} < 1$ and $K_{2m+1} > \frac12$. I wanted to prove these, but I could not just use induction (it's easy to see why if you try to prove $K_{2m} < 1$ by induction directly); but we know that, for large $m$, the last term which appears is very small - more specifically, the denominator of the last term is of order $O(m^2)$ - so that we can aim to create a sharper bound by adding in extra term(s).