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I need to determine if the following series cponverges. Which I believe I have done correctly, but then I must determine how many terms must be summed to guarantee an error no greater than $\frac {1}{10}$. This second part I am having some trouble with.

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^{0.001}}$. Clearly this series is alternating so I used the criteria for alternating series test.

First looking at the limit criteria as $a_n$ must go to $0$ for a alternating series to converge.

$lim \frac{1}{n^{0.001}} = 0$.

Then comparing the $n+1$ to $n$ we see that $\frac {1}{(n+1)^{0.001}}$ is clearly less than $\frac {1}{n^{0.001}}$. So this series must converge by the alternating series test.

Now looking at the second part I began to calculate the sum of the series,

$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^{0.001}} = 1- 0.993+ 0.9989- 0.9986$. Looking at this sum how can I possibly get it to be no greater than $\frac {1}{10}$

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    Have you learn about maximum error for an alternating series? By the way, you are going to need a lot of terms here. This series converges like a snail – 2017-01-24
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    @imranfat hahaha I know this is my problem – 2017-01-24
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    You essentially need to solve an equation involving the expression of the series and that $0.1$. Do you know what it is? – 2017-01-24
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    @imranfat no I am not quite sure – 2017-01-24
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    could you help me out with this? – 2017-01-24
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    http://tutorial.math.lamar.edu/Classes/CalcII/EstimatingSeries.aspx read bottom page... – 2017-01-24

1 Answers 1

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One may use the fact that $\frac1{n^{0.001}}$ converges monotonically to $0$, thus,

$$\left|\sum_{n=k}^\infty\frac{(-1)^{n+1}}{n^{0.001}}\right|<\frac1{k^{0.001}}$$

And just choose the proper value of $k$ (it's going to have to be big, as the series converges super slowly).

If one is interested in evaluating the series faster, applying an Euler transform, I get

$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^{0.001}}=\sum_{j=0}^\infty\frac1{2^{j+1}}\sum_{n=0}^j\binom jn\frac{(-1)^n}{(n+1)^{0.001}}$$

Estimating with an upper bound of $p$, I get

$$\begin{array}{c|c}p&\sum_{j=0}^p\frac1{2^{j+1}}\sum_{n=0}^j\binom jn\frac{(-1)^n}{(n+1)^{0.001}}\\\hline 1&0.500\ 173\ 226\ 752\ 386\ 869\ 519\\ 2&0.500\ 209\ 202\ 375\ 383\ 508\ 224\\ 3&0.500\ 219\ 829\ 114\ 274\ 006\ 399\\ 10&0.500\ 225\ 746\ 925\ 048\ 072\ 352\\ 20&0.500\ 225\ 760\ 827\ 952\ 674\ 479\\ 30&0.500\ 225\ 760\ 834\ 158\ 697\ 626\\ 40&0.500\ 225\ 760\ 834\ 162\ 478\ 372\end{array}$$

which is the correct value out $16$ places.

The exact value is given by

$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^{0.001}}=\eta(0.001)\\=0.500\ 225\ 760\ 834\ 162\ 481\ 008\ 261\ 982\ 830\ 758\ 637\ 859\ 937\ 556\ 152\ 849$$

which should agree conceptually with the linear approximation of the eta function around $x=0$:

$$\eta(x)\approx\left(\frac12\ln\frac\pi2\right)x+\frac12$$

$$\eta(0)\approx0.500225791$$

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    is that enough to show that the series converges? – 2017-01-24
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    or should I still use my AST – 2017-01-24
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    @hburt Convergence? You already showed convergence with AST, so I thought you wanted to calculate the sum out yourself with accuracy of $0.1$. – 2017-01-24
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    yes sorry my question was a bit confusing. Yes I do want to calculate the sum with accuracy of 0.1 – 2017-01-24
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    @hburt Yes. I mean, if you want to drop being correct to pick up "how to evaluate this much faster", than I can do that. Particularly, the solution is $\approx0.5$, for the record. – 2017-01-24
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    how did you calculate that? and how many terms must be summed to guarenetee that? – 2017-01-24
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    @hburt I bumped it to$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^{0.001}}\approx\lim_{x\to0^+}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n^x}=\eta(0)=0.5$$ – 2017-01-24
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    so how many terms is that? and is that alright to do to change it like that? it seems like that is the only reasonable thing to do – 2017-01-24
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    Be patient for a few moments. I'm trying to give you an answer XD – 2017-01-24
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    okay thanks so much – 2017-01-24
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    @hburt Phew! So many numbers! :D – 2017-01-24
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    haha that was a lot of work, thanks. So how many terms was that? – 2017-01-24
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    @hburt Uh.. I actually cheated and used a thing called a "series accelerator", which let's me approximate a series many folds faster. (i.e., I gave up on the original series. If you are wondering, the accelerator here used $p+1$ terms of the original series. In other words, approximation with only 11 terms) – 2017-01-24
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    okay and is that covered in your answer? – 2017-01-24
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    @hburt It is under "Euler transform", which is what I used. – 2017-01-24
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    thanks so much for your clear and precise answer – 2017-01-24
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    @hburt No problem :-) – 2017-01-24
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    @hburt I updated btw, in case you needed the first 30 or so digits. :P – 2017-01-24
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    thanks again for your time – 2017-01-24