If $S=1+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{3}{8}+...$ what is the closest integer to $e^S$

Question

If $S=1+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{3}{8}+...$ what is the closest integer to $e^S$ I thought that this series could be represented as $\ln{2}$ but it is $\ln{4}$ somehow? Any suggestions please send.

Answer 1

Starting from @vonbrand's answer, $$\begin{align*} S_n &= H_n - H_{\lfloor n / 4 \rfloor} \implies e^{S_n}=\exp\left( H_n - H_{\lfloor n / 4 \rfloor} \right)=\frac {\exp\left( H_n\right)}{\exp\left(H_{\lfloor n / 4 \rfloor}\right) } \end{align*}$$ Using the asymptotics of harmonic numbers, an approximation is $$e^{S_n}\sim\frac{n+\frac{1}{2}+\frac{1}{24 n}-\frac{1}{48 n^2} }{m+\frac{1}{2}+\frac{1}{24 m}-\frac{1}{48 m^2}}\qquad \text{where} \qquad m=\lfloor n / 4 \rfloor$$ Disregarding the floors, this would give for large values of $n$ $$e^{S_n}=4-\frac{6}{n}+\frac{19}{2 n^2}-\frac{39}{4 n^3}+O\left(\frac{1}{n^4}\right)$$

Answer 2

If your sum is really (there are certainly lots of other ways to go on!):

$\begin{align*} S_n &= 1 + \frac{1}{2} + \frac{1}{3} + \left(\frac{1}{4} - 1\right) + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \left(\frac{1}{8} - \frac{1}{2}\right) + \dotsm + \left(\frac{1}{12} - \frac{1}{3}\right) + \dotsm \\ &= \sum_{1 \le k \le n} \frac{1}{k} - \sum_{1 \le k \le \lfloor n / 4 \rfloor} \frac{1}{k} \end{align*}$

In terms of harmonic numbers:

$\begin{align*} H_n &= \sum_{1 \le k \le n} \frac{1}{k} \end{align*}$

your sum is just:

$\begin{align*} S_n &= H_n - H_{\lfloor n / 4 \rfloor} \end{align*}$

then, as hinted in comments, you can use the (rather crude) bound on the harmonic numbers $\ln n \le H_n \le \ln n + 1$:

$\begin{align*} \ln n - (\ln \lfloor n / 4 \rfloor + 1) &\le S_n \le \ln n + 1 - \ln \lfloor n / 4 \rfloor \end{align*}$

This tells you that, disregarding the floors, we have approximately:

$\begin{align*} \ln 4 - 1 &\le S_n \le \ln 4 + 1 \end{align*}$

Thus $e^{S_n}$ is between $4 / e = 1.4715$ and $4 e = 10.873$. Better bounds on $H_n$ give sharper values.