<s>[INST] 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 $e^S$ 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.
 [/INST] 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 $n$
$$e^{S_n}=4-\frac{6}{n}+\frac{19}{2 n^2}-\frac{39}{4
   n^3}+O\left(\frac{1}{n^4}\right)$$
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