Showing $\sum\limits_{n = 1}^∞\log(1+(-1)^{n-1}\frac{1}{n})$ is convergent

Question

Show that $$\sum_{n = 1}^\infty \log\left(1+(-1)^{n-1}\frac{1}{n}\right)$$ converges.

I want to say that the convergence/divergence of this series is equivalent to the convergence/divergence of $$\sum(-1)^{n-1}\frac{1}{n}.$$ Without the sign term I can show by L'hospital's rule that $$\lim_{n\to \infty}\frac{\log(1+1/n)}{1/n}=1.$$ But I don't know how to compare the given sries with $\sum(-1)^{n-1}/n$. Any suggestion is appreciated.

Answer 1

Note that $\log{(1 + x)} = x + O(x^2)$ in a neighborhood of $x = 0$.

So there exists $C$ such that $$-C x^2 < \log (1 + x) - x < C x^2$$

So the sum $$ \sum _ {n = 1} ^ \infty \left(\log {\left(1 + \frac{(-1)^n}n\right)} - \frac{(-1)^n}n\right)$$ converges absolutely. We conclude that the original sum converges too.

Answer 2

Grouping consecutive odd and even terms (that's allowed, because the terms tend to $0$ as $n\to\infty$, we get \begin{align}\sum\limits_{n = 1}^\infty\log\left(1+(-1)^{n-1}\frac{1}{n}\right)&=\sum\limits_{k = 1}^\infty\left(\log\left(1+\frac1{2k-1}\right)+\log\left(1-\frac1{2k}\right)\right)\\&=\sum\limits_{k = 1}^\infty\log\frac{2k}{2k-1}\frac{2k-1}{2k}=0\end{align} It means that all partial sums with an even number of terms are $0$, while the following partial sum with $2k+1$ terms is positive, but $<\dfrac1{2k+1}$.

Answer 3

Let $a_n := \prod_{k = 1}^n \left(1+(-1)^{k-1}\frac{1}{k}\right).\;$ Then $a_n = 1$ if $n$ is even and $1+\frac1n$ if $n$ is odd. Thus $a_n$ converges to $1$ and $\log(a_n)$ converges to $0$.

Answer 4

Just to add a note: Since $\ln(1+x)\leq x$ for all $x$, the convergence of

$$\sum_{n=1}^{\infty} \ln(1+x_n)$$

is, in general, implied by that of

$$\sum_{n=1}^{\infty} x_n.$$

(In fact, if $x_n>0$, it is possible to say more; since the convergence of the first sum is euivalent to that of $\prod_{n=1}^{\infty} (1+x_n)$ by definition, this is equal to $1+x_1+x_2+\cdots+(x_1x_2+x_1x_3+x_2x_3+\cdots)+\cdots$ which is strictly greater than $\sum_{n=1}^{\infty} x_n$, so the two convergences are actually equivalent.)

Answer 5

You can apply Leibniz's alternating series test directly to the given series. First notice that, if $n$ is even, then: \begin{align} \log\left(1+(-1)^{n-1}\frac 1n\right)=\log\left(\frac{n-1}{n}\right)=-\log\left(\frac{n}{n-1}\right)=-\log\left(1+\frac{1}{n-1}\right) \end{align} So: $$\sum_{n=1}^\infty \log\left(1+(-1)^{n-1}\frac 1n\right) = \sum_{n=1}^\infty (-1)^{n-1}a_n$$ where \begin{align} a_n= \begin{cases} \log(1+\frac{1}{n}) & \text{ if } & n \text{ is odd}\\ \log(1+\frac{1}{n-1}) & \text{ if } & n \text{ is even} \end{cases} \end{align} Clearly $a_n$ is nonnegative and $a_{n+1}\leq a_n$ and $\lim_n a_n=0$ so by Leibniz's test the original series converges.