Scrambled Summations Reupload

First, let us focus on $t$ . By revising our logarithm properties, we know that $\ln(10)\cdot\log(k) = \ln(k)$ because $\log_a(b)\cdot \log_b(c)= \log_a(c)$ , so $t$ can be rewritten as $\begin{aligned} t &= \sum_{k=1}^{n}\ln(10)\log(k) \\ &= \sum_{k=1}^{n}\ln(k) \\ &= \ln(n!) \end{aligned}$ Now, we can plug $t$ into our main calculation: $\begin{aligned} 1+\sum_{n=1}^{\infty}\frac{x^n(1-(n-1)!(-1)^n)}{n!} &= 1+\sum_{n=1}^{\infty}\frac{x^n-x^n(n-1)!(-1)^n)}{n!} \\ &= 1+\sum_{n=1}^{\infty}\frac{x^n}{n!}-\frac{x^n\sout{(n-1)!}(-1)^n}{n\sout{(n-1)!}} \\ &= 1+\sum_{n=1}^{\infty}\frac{x^n}{n!} - \sum_{m=1}^{\infty}\frac{x^m(-1)^m}{m} \\ &= \sum_{n=0}^{\infty}\frac{x^n}{n!} - \sum_{m=1}^{\infty}\frac{x^m(-1)^m}{m} \end{aligned}$ We know that the Maclaurin Series of $e^x$ is $\displaystyle \sum_{n=0}^{\infty}\frac{x^n}{n!}$ and for $\ln(1+x)$ is $\displaystyle - \sum_{m=1}^{\infty}\frac{x^m(-1)^m}{m}$ for $|x| < 1$ , so our main calculation shrinks down to $e^x + \ln(1+x)$

Now we can plug in $x=0.5$ and we see that the answer is $\large e^{0.5} + \ln(1.5) = \green{\boxed{2.05418637881\dots}}$