<s>[INST] Estimating Error due to replacing the sum $\sum\limits_{n=1}^{\infty} \frac{1}{n!} (\frac{1}{2})^n$ by the first $n$ terms Question: Estimating Error due to replacing the sum $\sum\limits_{n=1}^{\infty} \frac{1}{n!} (\frac{1}{2})^n$ by the first $n$ terms
All I can really say at this point is that the remainder $R_n$ when summing the first $n$ terms will be:
$$R_n = a_n [\frac{1}{2}\frac{1}{n+1} + (\frac{1}{2})^2\frac{1}{(n+1)^2} + (\frac{1}{2})^3\frac{1}{(n+1)^3} + ...]$$
Where $a_n = \frac{1}{n!} (\frac{1}{2})^n$. I can technically restrict $R_n$ by the following:
$$R_n < a_n[\frac{1}{2}\frac{1}{n+1} + (\frac{1}{2})^2 \frac{1}{(n+1)^2} + ...] = a_n\sum(\frac{1}{2(n+1)})^p$$
I am suppose to conclude that the error bounds $R_n < \frac{a_n}{2n+1}$, can anyone help me make that conclusion?
 [/INST] Note that we have
$$\begin{align}
\sum_{m=n+1}^\infty \frac{1}{m!2^m}&=\frac{1}{n!2^n}\sum_{m=n+1}^\infty \frac{n!2^n}{m!2^m}\\\\
&=\frac{1}{n!2^n}\sum_{m=n+1}^\infty \frac{n!}{m!2^{m-n}}\\\\
&=\frac{1}{n!2^n}\sum_{m=1}^\infty \frac{n!}{(m+n)!2^{m}}\\\\
&\le \frac1{n!2^n}\sum_{m=1}^\infty \frac1{(n+1)^m2^m}\\\\
&=\frac1{n!2^n}\frac{\frac1{2(n+1)}}{1-\frac1{2(n+1)}}\\\\
&=\frac1{(2n+1)n!2^n}
\end{align}$$
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