# smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$

If $S_n$ and $S_\infty$ are sums to $n$ terms and sum to infinity of a geometric progression $3,-\frac{3}{2},\frac{3}{4},...$ respectively, find the smallest integer value of $n$ such that $|S_n-S_\infty|<0.001$

My attempt,

$$|S_n-S_\infty|<0.001$$

$$|\frac{3[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}-\frac{3}{1-(-\frac{1}{2})}|<0.001$$

$$|2[1-(-\frac{1}2)^n]-2|<0.001$$

How to proceed? Thanks in advance.

• You are almost done ! – Khosrotash Aug 11 '17 at 14:49

$$|2[1-(-\frac{1}2)^n]-2|<0.001\\ |-2(\frac{-1}2)^n|<0.001\\ +2|(\frac{-1}2)^n|<0.001\\\to \text{abs function properties } |\frac{-1}{2^n}|=\frac{1}{2^n}\\ +2.\frac{1}{2^n}<\frac{1}{1000}\\ \frac{2^n}{2}>1000\\ 2^{n-1}>1000\\\text{note that } 2^{10}=\color{red} {1024>1000}\\\to \\n-1\geq 10\\n \geq 11$$

• Or I could use $(n-1)\ln2 >\ln1000$. Thanks a lot! – Mathxx Aug 11 '17 at 14:58

$$|\frac{3[1-(-\frac{1}{2})^n]}{1-(-\frac{1}{2})}-\frac{3}{1-(-\frac{1}{2})}|<0.001$$

$$|2[1-(-\frac{1}2)^n]-2| = |2(-\frac{1}2)^n|=\frac{1}{2^{n-1}}<0.001$$ ie $$-(n-1) \ln 2 < \ln (0.001) \implies n-1> \frac{\ln 1000 }{\ln 2}$$ hence the smalest integer is given by $$N=\left[\frac{\ln 1000 }{\ln 2} +1 \right] +1 = 11$$ $[x]$ stand for the floor of $x$.

For a convergent alternating sum $S_n=\sum\limits_{k=1}^na_k$, with $S=\lim\limits_{n\to\infty}S_n$, we have

\begin{align*} |S_n-S|&=\left|\sum_{k=1}^na_k-S\right|\\ &=\left|\sum_{k=n+1}^\infty a_k\right|\\ &\le\left|\sum_{k=n+1}^Na_k\right|&\text{for some }N>n+1\\ &=|a_{n+1}+\cdots+a_N|\\ &\le|a_{n+1}|+\cdots+|a_N|\\ &\le|a_{n+1}| \end{align*}

We have $a_k=3\left(-\frac12\right)^{k-1}$ for $k\in\mathbb N$. Thus

$$|S_n-S|\le\left|3\left(-\frac12\right)^n\right|=\frac3{2^n}$$

Then the approximate value of $S_n$ will fall within $0.001$ of $S$ for

$$\frac3{2^n}<0.001\implies2^n>3000$$

$2^{11}=2048$ and $2^{12}=4096$, so $n=12$ will guarantee $|S_n-S|<0.001$.

Unfortunately, as the other answers demonstrated, this isn't a hard limit. So perhaps this isn't the best method, but it gives a good starting point to find a better $n$.

• Thanks for posting a general method for this type of question! – Mathxx Aug 12 '17 at 3:40