1. ## series

Is there a better way to do this problem?

For the following convergent series, at least how many terms do you have to sum so that the partial sum is within 0.001 of the sum of the series?

$\displaystyle 1 - \frac{2}{3}+ \frac{3}{9} - \frac{4}{27}+...+(-1)^{k-1} \frac{k}{3^{k-1}}+...$

If you add the first k terms, the remainder is

$\displaystyle (-1)^{k} \frac{k+1}{3^{k}} + (-1)^{k+1} \frac{k+2}{3^{k+1}} + (-1)^{k+2} \frac{k+3}{3^{k+2}}+... = \ (-1)^{k} \frac{1}{3^{k}} \Big((k+1)-\frac{(k+2)}{3} $$\displaystyle + \frac{k+3}{9} +...\Big) Using the fact that \displaystyle x^{k+1} - \frac{x^{k+2}}{3} +\frac{x^{k+3}}{9}+... = \frac{x^{k+1}}{1+\frac{x}{3}}\ , \ |\frac{x}{3}| < 1 take the derivative of both sides \displaystyle (k+1)x^{k}- \frac{k+2}{3}x^{k+1} + \frac{k+3}{9}x^{k+2} + ... = \frac{(k+1)x^{k}(1+\frac{x}{3})-\frac{1}{3}x^{k+1}}{(1+\frac{x}{3})^{2}} let x=1 \displaystyle (k+1) - \frac{k+2}{3} + \frac{k+3}{9} + ... = \frac{12k+9}{16} so we want \displaystyle \Big|(-1)^{k}\frac{1}{3^{k}}\frac{12k+9}{16}\Big|<\frac{1 }{1000} \displaystyle \frac{12k+9}{3^{k}} < \frac{2}{125} but that can't be solved explicitly for k although it appears to be true for k >8 2. See attachment 3. Hello, Random Variable! I can help get you started . . . A least how many terms do you have to sum so that the partial sum is within 0.001 of the sum of the series? \displaystyle 1 - \frac{2}{3}+ \frac{3}{9} - \frac{4}{27}+ \hdots + \frac{(\text{-}1)^{k-1}k}{3^{k-1}}+ \hdots \displaystyle \begin{array}{ccccc}\text{We have:} & S &=& 1 - \dfrac{2}{3} - \dfrac{3}{9} + \dfrac{4}{27} - \dfrac{5}{81} + \hdots \\ \\[-3mm] \text{Multiply by }\frac{1}{3}: & \dfrac{1}{3}S &=& \quad\;\;\; \dfrac{1}{3} - \dfrac{2}{9} + \dfrac{3}{3^3} - \dfrac{4}{3^4} + \hdots\end{array} \displaystyle \text{Subtract: }\qquad\quad\frac{2}{3}S \;\:=\;\;\underbrace{1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} - \frac{1}{81} + \hdots}_{\text{geometric series}} The geometric series has its sum: .\displaystyle \frac{1}{1-\left(\text{-}\frac{1}{3}\right)} \:=\:\frac{1}{\frac{4}{3}} \:=\:\frac{3}{4} Hence, we have: .\displaystyle \frac{2}{3}S \:=\:\frac{3}{4} \quad\Rightarrow\quad S \:=\:\frac{9}{8} \quad\Leftarrow\text{ sum of the series} 4. Originally Posted by Calculus26 See attachment I might be wrong, but it seems like what you did is find the first term whose absolute value is less than 0.001. 5. Yes that is from the theorem on the sum of an alternating series: |S-Sn|<a(n+1) the difference between the actual sum S of an alternating series and the nth partial sum Sn is bounded by the n+1st term of the sequence an. 6. If interested I have the proof of this Theorem on my web site Infinie Series 7. Originally Posted by Calculus26 Yes that is from the theorem on the sum of an alternating series: |S-Sn|<a(n+1) the difference between the actual sum S of an alternating series and the nth partial sum Sn is bounded by the n+1st term of the sequence an. But what if you couldn't use that fact? 8. . 9. But what if you couldn't use that fact? Well its a handy tool and I always try to look classify a series and determine the best approach. Soroban cleverly worked it into a geometric series-- something I wish I had thought of. 10. Originally Posted by Soroban Hello, Random Variable! I can help get you started . . . \displaystyle \begin{array}{ccccc}\text{We have:} & S &=& 1 - \dfrac{2}{3} - \dfrac{3}{9} + \dfrac{4}{27} - \dfrac{5}{81} + \hdots \\ \\[-3mm]$$\displaystyle \text{Multiply by }\frac{1}{3}: & \dfrac{1}{3}S &=& \quad\;\;\; \dfrac{1}{3} - \dfrac{2}{9} + \dfrac{3}{3^3} - \dfrac{4}{3^4} + \hdots\end{array}$

$\displaystyle \text{Subtract: }\qquad\quad\frac{2}{3}S \;\:=\;\;\underbrace{1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} - \frac{1}{81} + \hdots}_{\text{geometric series}}$

The geometric series has its sum: .$\displaystyle \frac{1}{1-\left(\text{-}\frac{1}{3}\right)} \:=\:\frac{1}{\frac{4}{3}} \:=\:\frac{3}{4}$

Hence, we have: .$\displaystyle \frac{2}{3}S \:=\:\frac{3}{4} \quad\Rightarrow\quad S \:=\:\frac{9}{8} \quad\Leftarrow\text{ sum of the series}$

It looks like you made a sign error. I think the sum is 9/16.

This is what I did:

$\displaystyle x -\frac{x^{2}}{3} + \frac{x^{3}}{9} - \frac{x^{4}}{27} + ... = \frac{x}{1+\frac{x}{3}} \ , \ \Big|\frac{x}{3}\Big|<1$

differentiate

$\displaystyle 1 - \frac{2x}{3} + \frac{3x^{2}}{9} - \frac{4x^{3}}{27} + ...= \frac{9}{(3+x)^{2}}$

let x = 1

$\displaystyle 1 -\frac{2}{3} +\frac{3}{9}-\frac{4}{27} + ... = \frac{9}{16}$

so know I need to solve $\displaystyle \Big|\frac{9}{16} - (1-\frac{2}{3} + \frac{3}{9} - \frac{4}{27}+ ...(-1)^{k}\frac{k}{3^{k-1}})\Big| < 0.001$ for k ?

That difference is the remainder.