I have a sum to calculate:
$$\sum_{k=51}^{\infty} \frac{(-1)^{k-1}}{k} $$
And I have no idea about how to proceed. What kind of techniques are available to calculate this?
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The parentheses around the $-1$ are of utter importance; else the series will be divergent. – AlexR Oct 05 '14 at 22:02
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What do you know about $$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k}$$ – graydad Oct 05 '14 at 22:02
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@graydad I know that I can use a taylor expansion to get it to be -ln(1+1) – iveqy Oct 05 '14 at 22:05
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Right! Well, almost... There shouldn't be a negative out front of the $\ln$ should there? – graydad Oct 05 '14 at 22:07
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@graydad true it shouldn't. But how does that help me? I can state it as $ln(2) - \sum_{k=1}^{50} \frac{(-1)^{k-1}}{k}$ but how to I calculate that sum? – iveqy Oct 05 '14 at 22:10
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You nailed it. Do you have access to a computer where you can calculate $\sum_{k=1}^{50} \frac{(-1)^{k-1}}{k}$? Because there is no quick way to calculate that by hand. I guarantee that is an ugly number. – graydad Oct 05 '14 at 22:13
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yes, but shouldn't it be possible to do it without a computer? The result itself is not as interesting as learning... – iveqy Oct 05 '14 at 22:14
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It is definitely possible to solve by hand. It's just that you'll have to find a common denominator for the fifty fractions that you have to add together. I recommend using a denominator of $50$ factorial. I think the equation $$\sum_{k=51}^\infty \frac{(-1)^{k-1}}{k} = \ln(2)-\sum_{k=1}^{50} \frac{(-1)^{k-1}}{k} $$ is exactly the answer you are looking for. I should clarify about when I said "possible to solve by hand." You do NOT want to do this by hand. It will take way too long and is just an exercise in adding fractions. – graydad Oct 05 '14 at 22:17
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And how to do this? – iveqy Oct 05 '14 at 22:20
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1Possible duplicate of The sum $1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots-(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots)$ does not exist. – Mohan Sharma Jan 15 '18 at 08:45
1 Answers
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Start with the Taylor series of $\log(1+x)$:
$$\log(1+x) = \sum_{k=1}^\infty\frac{(-1)^{k-1}x^k}{k}$$
and take $x=1$ (that this is allowed follows from Abel's theorem) to find
$$\log(2) = \sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$$
Now split the sum into two parts:
$$\log 2 = \sum_{k=1}^\infty\frac{(-1)^{k-1}}{k} = \sum_{k=1}^{50}\frac{(-1)^{k-1}}{k} + \sum_{k=51}^\infty\frac{(-1)^{k-1}}{k}$$
The first term can be calculated on a computer (there is no known (useful) formula you can use so a calculation is needed here)
$$ \sum_{k=1}^{50}\frac{(-1)^{k-1}}{k} = \frac{2117413358024481287753}{3099044504245996706400} \approx 0.68324716$$
Thus giving us
$$\sum_{k=51}^\infty\frac{(-1)^{k-1}}{k} = \log 2 - \frac{2117413358024481287753}{3099044504245996706400} \approx 0.00990002$$
If you are only looking for an approximate value: The terms in the series $\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}$ alternates being positive and negative and the absolute value of the terms decreases with $k$. This means that the sum you are after satsify:
$$\left|\sum_{k=51}^\infty\frac{(-1)^{k-1}}{k}\right| = \left|\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k} - \sum_{k=1}^{50}\frac{(-1)^{k-1}}{k}\right| < \left|\frac{(-1)^{51-1}}{51}\right| = \frac{1}{51}$$
so the sum you are after is less than $0.02$ (which is indeed the case).
Winther
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