Proof that $0< e-\ln 2-2 < \frac 3 {100}$

You may have learnt the estimates $e\approx 2.7$ and $\ln 2 \approx 0.7$, thus yielding $e-\ln 2 \approx 2$. A calculator indicates more precisely that $e-\ln 2\approx 2.025$.
Out of curiosity I'd like to see self-contained (requiring only pen and paper) proofs that $$0< e-\ln 2-2 < \frac 3 {100}.$$

Here's my attempt. Surely by power series expansion, $$e-\ln 2-2 = -\frac 16+\sum_{k=4}^\infty (-1)^k\left(\frac 1{k}+\frac{(-1)^k}{k!}\right),$$and it is easily verified that the series in the RHS is alternating.Consequently we obtain the bound$a_n \leq e-\ln 2-2 \leq b_n$ for each $n\geq 4$, where$$a_n = -\frac 16 +\sum_{k=4}^{2n+1} (-1)^k\left(\frac 1{k}+\frac{(-1)^k}{k!}\right)$$and$$b_n = -\frac 16 +\sum_{k=4}^{2n} (-1)^k\left(\frac 1{k}+\frac{(-1)^k}{k!}\right).$$

A computer indicates that $a_{10}>0$ and $b_{52}<\frac 3 {100}$, thus yielding a proof, which is not satisfactory since it requires a computer/calculator.