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Estimate the sum correct to three decimal places : $$\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$$

This problem is in my homework. I find that n = 22 when use Maple to solve this. (with some programming) But, in my homework, teacher said find the formula for this problem.

Thanks :)

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    Does that sum really start at $n=0$?2012-06-10

5 Answers 5

10

By the Alternating Series Test, the error to an alternating series with monotonically decreasing terms is the next term to be added. Thus, to get three decimal places, we would need to find an $n$ so that $n^3>2000$, which would be $n=13$. Thus, summing the first 12 terms should get you to within 3 decimal places.

  • 2
    To get 3 decimal places accuracy, don't we need $n^3> 10,000$? (And the cube root of 10,000 is indeed about 21.5.)2012-06-10
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    Three decimal places would mean an error of less than $.0005=\frac{1}{2000}$.2012-06-10
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    In practice here even 22 terms is not enough to get the correct result (rounded!) since $s(22)=-0.90149891\cdots$ and $s(23)=-0.90158110\cdots$ (23 is fine since $s(24)=-0.901508\cdots$ !).2012-06-10
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    @RaymondManzoni: That is true. To make sure the first $3$ decimal places are correct, it could require many more places to be computed to make sure that rounding and subsequent terms don't affect the first three digits. That is why I consider "to $3$ places" to mean "an error of less than $.0005$" (and this is simply my opinion).2012-06-10
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    @robjohn: I understand your point of view (mine would be similar) as well as Ragib's. It was just fun to notice that finding the number of terms required was more prickly than it seemed...2012-06-10
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    An interesting follow-up that occurs: how many terms do you need to sum in $\sum_{n=0}^\infty \frac{3(-1)^n}{2^{n+1}}$ to even find out what the *first* decimal place is?2012-06-10
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    @benmachine: in fact 4 ($s(4)=-0.46875,\ s(5)=-0.515625\ $ to return $-0.5\ $ he he) but I see your point if you want the nearest integer. Anyway who would evaluate this numerically here? :-)2012-06-10
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    @RaymondManzoni: hm, are you sure? I think the first four terms are $3/2$, $-3/4$, $3/8$, $-3/16$, and they add up to $-15/16$, and if you include the next term they add up to $33/32$, so the first decimal place flips from 9 to 0. I was trying to make a series that oscillates around 1, to make the point that there's no guarantee you can get *any* of the expansion just by summing finite prefixes...2012-06-10
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    @ben: Oups I forgot the $n=0$ so that the number of terms was 5 ($\sum_0^4 = 1.03125,\ \sum_0^5 = 0.984375$) to get the result (rounded to 1 decimal!) of $1.0$ (I supposed too that we were 'rounding' not truncating). Anyway your example is neat,2012-06-10
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    Oh, yeah, I'm silly. But you could take half of that sequence, and then it'd oscillate around 0.5, so you couldn't say whether the nearest integer was 0 or 1, which is I guess what you meant before.2012-06-10
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    @benmachine: yes without the $n=0$ term (minus) your example was better!2012-06-10
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    @benmachine: *with* the $n=0$ term, there are some problems :-)2012-06-10
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    Averaging two consecutive partial sums reduces the needed number of terms. Since the sum is in $(s_{n-1},s_{n})$, we would allow $s_{n}-s_{n-1}\leq0.0005\cdot2$ to ensure that $\frac{s_n+s_{n-1}}{2}$ is within $0.0005$ of the sum. Thus we'd have $\frac{1}{n^3}\leq0.001$, implying $n$ could be as small as $10$ if we average the 9th and 10th partial sums.2014-06-07
4

For alternating sums $\sum(-1)^n a_n$ with $a_n> 0 $ strictly decreasing there is a simple means to estimate the remainder $\sum^\infty_{k=n} (-1)^k a_k$. You can just use $a_{n-1}$.

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$$\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^3} = \sum_{k=1}^{\infty} \left(\dfrac{1}{k^3}-\dfrac{1}{(k+1)^3}\right)= \sum_{n=1}^{\infty} \left(\dfrac{(k+1)^3-k^3}{k^3(k+1)^3}\right)$$

$$= \sum_{k=1}^{\infty} \dfrac{(k+1-k)(k^2+k+1)}{k^3(k+1)^3}$$

$$= \sum_{k=1}^{\infty} \dfrac{k^2+k+1}{k^3(k+1)^3}$$

$ $

Note that for the sum to be accurate within 3 decimal planes the nth term must be less than 0.001

Therefore we have $$\dfrac{k^2+k+1}{k^3(k+1)^3} < \dfrac{1}{1000}$$

You will notice that answer can either be $2k$ or $2k+1$

I will think about it later, how to pin point it, I will have to go now, let me know what you guys think about it

Using wolfram you will find that $k=5.18$ satisfies the inequality

Therefore the smallest value of $k$ to satisfy the inequality will be $6$

And hence the answer can either be $12$ or $13$

In fact it should be $n=13$ terms, do you need me to explain that? Try thinking first, there is a clear logical reason for $n = 13$ and not $n=12$.

1

With a little bit of approximation, we can achieve the desired $5\times 10^{-4}$ accuracy with fewer terms.

Notice

$$\sum_{n=1}^\infty \frac{(-1)^n}{n^3} = -\left[\sum_{n=1}^{\infty}\frac{1}{n^3}- \sum_{n=1}^\infty\frac{2}{(2n)^3}\right] = -\frac34\sum_{n=1}^\infty \frac{1}{n^3} = -\frac34\zeta(3)$$

If we want to estimate the sum at the left accurate up to $5\times 10^{-4}$, it just suffices to estimate $\zeta(3)$ accurate up to $6.67\times 10^{-4}$. In the sum of $\zeta(3)$, if we pick a $m$ and replace $\displaystyle\;\frac{1}{n^3}$ by $\displaystyle\;\frac{1}{n^3-n}$ for $n \ge m$, the error introduced $\mathcal{E}_m$ is

$$\mathcal{E}_m =_{def} \sum_{n=m}^\infty\left(\frac{1}{n^3-n} - \frac{1}{n^3}\right) = \sum_{n=m}^\infty \frac{1}{(n^2-1)n^3} \le \frac{m^2}{m^2-1}\sum_{n=m}^\infty \frac{1}{n^5} $$ Since $\displaystyle\;\frac{1}{x^5}$ is a convex function, we have

$$\frac{1}{n^5} \le \int_{n-1/2}^{n+1/2} \frac{dx}{x^5} \quad\implies\quad \mathcal{E}_m \le \frac{m^2}{m^2-1}\int_{m-1/2}^\infty \frac{dx}{x^5} = \frac{m^2}{4(m^2-1)(m-1/2)^4} $$ For $m = 5$, we have $\displaystyle\;\mathcal{E}_5 = \frac{25}{39366} \approx 6.35\times 10^{-4}\;$. This is already good enough for our purposes. Notice

$$\sum_{n=m}^{\infty}\frac{1}{n^3-n} = \sum_{n=m}^{\infty}\frac{1}{(n-1)n(n+1)} = \sum_{n=m}^{\infty}\left(\frac{1}{2(n-1)n}-\frac{1}{2n(n+1)}\right) = \frac{1}{2(m-1)m} $$ We find the original sum is approximately equal to $$-\frac34 \left( 1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{2\cdot 4\cdot5} \right) = -\frac{10391}{11520} \approx -0.90199653$$ with an error smaller than $5\times 10^{-4}$. Compare this with the exact value of the sum $\approx -0.90154268$, the difference $\approx -4.5385 \times 10^{-4}$ is indeed smaller than $5 \times 10^{-4}$.

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Averaging the 9th and 10th partial sums will do it, as in robjohn's answer and my comment there. Just for fun, as an alternative to alternating series methods, you could group consecutive terms to give a positive series and then use integral approximation.

$$\begin{align} S:=\sum_{n=1}^\infty\frac{(-1)^n}{n^3} &=-1+\sum_{n=1}^\infty\frac{1}{8n^3}-\frac{1}{(2n+1)^3}=-1+\sum_{n=1}^\infty f(n)\\ \end{align} $$

And then $$\begin{align} -1+\sum_{n=1}^{N}\frac{1}{8n^3}-\frac{1}{(2n+1)^3}+\int_{N}^\infty\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx

The difference between the outer bounds is $$\begin{align} \int_{N-1}^{N}\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx &=\left[{-{\frac1{16x^2}}}+\frac{1}{4(2x+1)^2}\right]_{N-1}^{N}\\ &={-{\frac1{16N^2}}}+\frac{1}{4(2N+1)^2}+{{\frac1{16(N-1)^2}}}-\frac{1}{4(2N-1)^2}\\ \end{align}$$ We'd like this difference to be less than $0.001$. This way we could average the two outer bounds as an estimate for the sum and know that our estimate was within $0.0005$. This happens for $N=5$. So an estimate that will work out to within $0.0005$ is

$$\sum_{n=1}^{11}\frac{(-1)^n}{n^3}+\frac12\int_4^{5}\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx$$

which works out to

$$\sum_{n=1}^{11}\frac{(-1)^n}{n^3}+\frac{24209}{125452800}\approx-0.9016748\ldots$$