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$$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$ I personally have such an idea: try to make geometric series like this $$1-\frac{1}{\sqrt{10}}(1-\frac{1}{10}+\frac{1}{100}+\cdot\cdot\cdot)-\frac{1}{10}(1+\frac{1}{10}+\cdot\cdot\cdot)$$ using 3 5 7 terms and 2 4 6 terms to form two geometric series. then if we calculate sum for them we will have $$S_1 = \frac{10}{11}$$ and $$S_2 = \frac{10}{9}$$ and after substituting them we'll finally have sum of series $$S = \frac{88-9\sqrt{10}}{99}$$ Is everything right here? But how to check it for convergence?

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  • $\begingroup$ Please check your + - - + - + $\endgroup$ May 20, 2017 at 21:49
  • $\begingroup$ @Salahamam_Fatima everything is correct here $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$ $\endgroup$
    – J.Doe
    May 20, 2017 at 21:50
  • $\begingroup$ Compare your series to the absolutely convergent geometric series with common ratio $r=\frac{1}{\sqrt{10}}$. $\endgroup$
    – sharding4
    May 20, 2017 at 21:53
  • $\begingroup$ I would consider three geometric series, one with all terms positive ( 1,4,7,...) , two with all terms negative (2,5,8... and 3,6,9,... ) and ratio for all of them $\frac{1}{10\sqrt{10}}$. $\endgroup$ May 20, 2017 at 22:02
  • $\begingroup$ @J.Doe Mmm why your result is different from mine? :\ $\endgroup$
    – Ixion
    May 20, 2017 at 22:24

4 Answers 4

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Write your sum like: $$S = {S}_1 + {S}_2 + {S}_3$$ Where, $${S}_1 = 1 + \frac{1}{\sqrt{10}\cdot 10} + \frac{1}{10^{3}} + ... = \frac{1}{1 - \frac{1}{10 \cdot\sqrt(10)}} = \frac{10\sqrt{10}}{10\sqrt{10} - 1}$$

$${S}_2 = -\frac{1}{\sqrt{10}} - \frac{1}{10^{2}} - ... = -\frac{1}{\sqrt{10}} \cdot{S}_1$$

$${S}_3 = -\frac{1}{10} - \frac{1}{10^{2}\sqrt{10}} - ... = -\frac{1}{10}\cdot{S}_1$$

And finally:

$$S = \frac{9\sqrt{10} - 10}{10\sqrt{10} - 1}$$

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My idea: suppose that $$1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}-\frac{1}{10^{2}}-\frac{1}{10^{2}\sqrt{10}}+\cdot\cdot\cdot$$ is a convergent serie. Let $S$ the sum of the serie $$S= 1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}\left(1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\cdot\cdot\cdot\right)=1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}S$$

The sum of the serie must satisfy the equation

$$S= 1-\frac{1}{\sqrt{10}}-\frac{1}{10}+\frac{1}{10\sqrt{10}}S$$

i.e. $$S=\frac{10\sqrt{10}-90}{\sqrt{10}-100}$$

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The identity of a series is often ambiguous when only a few example terms are given without a general rule for generating the rest. I assume that the general rule in this case is that the magnitude of each term is $\newcommand{r}{{\sqrt{10}}} 1/\r$ times the magnitude of the previous term, while the signs repeat every three terms: one positive term followed by two negative terms.

If you try to split the series after the first into two subseries by taking alternating terms (terms $2,4,6,\ldots$ in one subseries and $3,5,7,\ldots$ in the other), you get this: \begin{multline} 1 - \frac1\r \left(1 - \frac1{10} + \frac1{10^2} + \frac1{10^3} - \frac1{10^4} + \frac1{10^5} + \frac1{10^6} - \frac1{10^7} + \frac1{10^8} +-\cdots \right)\\ - \frac1{10}\left(1 + \frac1{10} - \frac1{10^2} + \frac1{10^3} + \frac1{10^4} - \frac1{10^5} + \frac1{10^6} + \frac1{10^7} - \frac1{10^8} +-\cdots \right)\\ \end{multline}

Although the magnitudes of each subseries form a geometric series with ratio $1/10,$ the subseries are not geometric with ratio $1/10,$ because the signs are not all the same. Just as in the original series, the signs change in patterns that repeat every three terms: $+,-,+$ for the first subseries, $+,+,-$ for the second. So the method proposed in the question is not correct.

In a geometric series with a positive ratio, all terms have the same sign. In a geometric series with a negative ratio, the terms have alternating signs, for example $+,-,+,-,+,-,\ldots.$ There is no ratio that will produce both positive an negative signs with a pattern that repeats every three terms.

There are various ways to deal with this by rearranging the series, just not in exactly the way attempted in the question. Of course, before we start rearranging the series we should make sure that the rearrangement would not affect the convergence or sum of the series. According to the ratio test, the series is absolutely convergent, and therefore it is OK to split the series into subseries or to rearrange the terms in other ways.

One rearrangement that works is to separate the series into three subseries as shown in another answer.

Another possible rearrangement is to combine the terms in groups of three: $$ \left(1-\frac1\r-\frac1{10}\right) + \left(\frac1{10\r}-\frac1{10^2}-\frac1{10^2\r}\right) + \left(\frac1{10^3}-\frac1{10^3\r}-\frac1{10^4}\right) + \cdots. $$

This is a geometric series with ratio $\frac1{10\r}$ and first term $\left(1-\frac1\r-\frac1{10}\right),$ which you can evaluate using the usual methods for a geometric series.

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To assume that a series is convergent to compute the value is not valid Ixion. i.e. $S=1+2+4+6+8+...$ is obviously not convergent although $S=1+2(1+2+3+4+...)=1+2S$ would imply $S=-1$.

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  • $\begingroup$ In your example $1+2+4+6+8+$... is not convergent. $\endgroup$
    – Ixion
    May 20, 2017 at 22:13
  • $\begingroup$ Oh, I got the message now. Sorry, english is not my mother language. I think that the method I used is correct if I assume that the serie is convergent. $\endgroup$
    – Ixion
    May 20, 2017 at 22:17
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    $\begingroup$ $S = 1 + 2\ \overbrace{\left(\,1 + 2 + 3 + 4 + \cdots\,\right)}^{\displaystyle\not=\ S}$. $\endgroup$ May 21, 2017 at 5:03

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