evaluate the sum of an alternating harmonic series with a fixed limit

Question

Hi I stumbled across an alternating harmonic sum with a fixed limit on an practice exam and I've no idea how to calculate this sum. \begin{equation} \sum_{1}^{100}\frac{(-1)^{k-1}}{k}. \tag{1} \end{equation} Now the point of the exercise is to make an approximation to this sum and compare it to $$\sum_{1}^{\infty}\frac{(-1)^{k-1}}{k} \tag{2}.$$ So naturally I tried to convert $(1)$ into an integral so I could make an approximation based on that, however I found it hard to integrate $$\frac{\cos(x)}{x}. $$ Then I tried to make use of Leibniz criterion but that got me nowhere. Since I guess there is a way to make an approximation of $(1)$. I would gladly accept any tip on how to solve this! Thanks for reading! :)

Answer 1

Here is an easier way to prove the answer is log(2).

$$\ln\left(\dfrac{1}{1-x}\right) + C = \int \dfrac{1}{1-x} dx = \sum_{k=1}^{\infty} \int x^{k-1} dx = \sum_{k=1}^{\infty} \dfrac{x^{k}}{k} $$

Assuming $C=0$ (which can be shown choosing x=0), at $x=-1$ we get

$$\ln\left(\dfrac{1}{2}\right) = \sum_{k=1}^{\infty} \dfrac{(-1)^{k}}{k}$$

For your sum, a well known estimate of an alternating series $\sum(-1)^nb_n$ where $\{b_n\} \subset \mathbb{R}_+$ is given by

$$|S-S_N| \le b_{N+1}$$

Since $S = \ln(2)$, we have

$$\left|\ln(2) - \sum_{n=1}^{100}(-1)^{n-1}\cdot \dfrac{1}{n}\right| \le \dfrac{1}{100+1}$$

i.e.

$$\ln(2)-\dfrac{1}{101} \le \sum_{n=1}^{100}(-1)^{n-1}\cdot \dfrac{1}{n} \le \ln(2)+\dfrac{1}{101}$$

Answer 2

Here's one way to think about it...

For $n$ even, \begin{align*} \sum_{k=1}^{n+1}\frac{\left(-1\right)^{k-1}}{k} & =\sum_{k=0}^{n}\frac{\left(-1\right)^{k}}{k+1}\\ & =\sum_{k=0}^{n/2}\frac{1}{2k+1}-\sum_{k=0}^{n/2}\frac{1}{2k+2}\\ & =\frac{1}{2}\left(H_{\frac{n+1}{2}}+\log4\right)-\frac{1}{2}\left(H_{\frac{n}{2}+1}\right)\\ & =\frac{1}{2}\left(H_{\frac{n+1}{2}}-H_{\frac{n}{2}+1}\right)+\log2 \end{align*} where $H_{j}$ is the $j$-th harmonic number. As $n\rightarrow\infty,$ $H_{\frac{n}{2}}-H_{\frac{n+1}{2}}\rightarrow0$, and hence we are left with $\log2$.

Answer 3

Starting from par's answer, you can use the expansion of the harmonic numbers for large values and, in such a case, you find that, for even values of $n$, $$\begin{equation} S=\sum_{k=1}^{2m}\frac{(-1)^{k-1}}{k}=\log (2)-\frac{1}{4 m}+\frac{1}{16 m^2}-\frac{1}{128 m^4}+O\left(\left(\frac{1}{m}\right)^5\right) \end{equation}$$ Using this expansion, for $m=50$, you will find $$S\approx \log (2)-\frac{3980001}{800000000} =0.68817217930994530942$$ while the exact value would be $$S=\frac{47979622564155786918478609039662898122617}{697203752297124771645338089353123035 56800} \approx 0.68817217931019520324$$