<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] 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.
</s>