The sum: $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$ using Riemann Integral and other methods

Question

I need to prove the following: $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+(-1)^{n+1}\frac{1}{n}+\cdots=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}=\ln(2)$$

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Method 1:

The series $\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$ is an alternating series, thus it is convergent, say to $l$. Therefore, both $s_{2n}$ and $s_n$ are convergent to the same limit $l$.

$$ \begin{align} s_{2n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2n} & =\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}\right) \\[10pt] & =\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \end{align} $$

It is an easy exercise to prove that

$$\lim_{n \to \infty }s_{2n}=\lim_{n \to \infty }s_n =\lim_{n \to \infty }\left [ \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n} \right ]=\ln(2)$$ which implies that the given alternating series converges to $l=\ln 2$.

However, I am interested to see proof of this problem using the definition of the Riemann Integral as a sum of infinitely many rectangles of widths tending to zero. I tried to come up with proof for this, but I couldn't. Can anyone share, please?

Also, I am interested to see other methods of solving this problem (other than my method and the Riemann method.) If anyone of you is aware of any other methods, please share.

Answer 1

Usually, to use the Riemann integral, alternating terms cause a problem. We want to have the finer partitions converge nicely, but an alternating series does not allow this. So, as far as I can see, we pretty much have to use the $\zeta$ trick you employ in your method: $$ \begin{align} &\frac11-\frac12+\frac13-\frac14+\dots+\frac1{2n-1}-\frac1{2n}\\ &=\left(\frac11+\frac12+\dots+\frac1{2n}\right)-2\left(\frac12+\frac14+\dots+\frac1{2n}\right)\\ &=\frac1{n+1}+\frac1{n+2}+\frac1{n+3}+\dots+\frac1{2n}\\ &=\sum_{k=n+1}^{2n}\frac{n}{k}\frac1n\tag{1} \end{align} $$ Then to use $(1)$ as a Riemann sum (with $x=k/n$ and $\mathrm{d}x=1/n$) for $$ \int_1^2\frac1x\,\mathrm{d}x=\log(2)\tag{2} $$

Answer 2

Here's another method by the Riemann integral, but not by definition: $$ \sum_{n=1}^{\infty }(-1)^{n+1}\frac{1}{n}= \lim_{m\to\infty}\sum_{n=1}^{m}(-1)^{n+1}\frac{1}{n}= $$ $$ \lim_{m\to\infty}\int_0^1(1-x+\ldots+(-1)^{m-1}x^{m-1})\,dx= \lim_{m\to\infty}\int_0^1\frac{1-(-x)^m}{1+x}\,dx= \int_0^1\frac{dx}{1+x}=\ln2. $$

Answer 3

You can use the following geometric series $$\sum_{n=0}^\infty x^n=\frac{1}{1-x},|x|<1.$$ Then integrate both sides to get $$\sum_{n=1}^\infty \frac{x^n}{n}=\int_0^x\frac{1}{1-t}dt=-\ln (1-x),|x|<1.$$ Since the LHS converges at $x=-1$, letting $x\to-1$ will give us $$\sum_{n=1}(-1)^{n+1}\frac{1}{n}=\ln2.$$

Answer 4

As you asked for other methods of solving this problem, here is one using Taylor series.

Now that you know you are looking for the RHS of $\ln(2)$, expand $f(x) = \ln(x)$ into Taylor series around $x=1$:

$f^{(n)}(x) = (-1)^{n+1} x^{-n} (n-1)!,$ for $n>0, x \neq 0$

so choosing to evaluate at $x=1$, using $f(1) = \ln(1) = 0$, we get

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

and using $x=2$, $\ln 2 = \sum_{n=1}^\infty (-1)^{n+1}/n$ as desired.

Answer 5

However one should note that, $$\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$$
is not absolutely convergent as $$\sum_{n=1}^\infty |(-1)^{n+1}\frac{1}{n}|=\sum_{n=1}^\infty \frac{1}{n}$$ is divergent. (Such a series is said to be conditionally convergent)

Hence there is no fixed value for the series. (Refer to the Riemann Series Theorem) $\ln 2$ is just a value that the series would take for the particular permutation of $S_{2n}$.