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Does $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converge? I used the integral test and the function $f(x)=\frac{sin (\frac {\pi}{2}x)}{x}$ to determine that the series converges. I wanted to know if this function $f(x)$ is a valid choice to use the integral test on.

Philip L
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  • Check this: https://math.stackexchange.com/q/716/42969 – Martin R Mar 15 '21 at 17:14
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    The integral test is for decreasing sequences, it cannot be used here. – Martin R Mar 15 '21 at 17:14
  • @PhilipL I assume you are not used to Leibniz's test. I have edited my answer, check it out. – CHAMSI Mar 15 '21 at 17:25
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    Write

    $$\begin{align} \sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n}&=\sum_{n=1}^N \left(\frac1{2n-1}-\frac1{2n}\right)\\ &=\sum_{n=1}^N \frac1{2n(2n-1)}\\ &\le \frac12 +\sum_{n=2}^N \frac1{2n(2n-2)}\\ &=\frac12+\frac14 \sum_{n=2}^N \left(\frac1{n-1}-\frac1n\right)\\ &=\frac12+\frac14 \left(1-\frac1N\right)\\&\le \frac34 \end{align}$$

    and conclude the series converges.

     – Mark Viola Mar 15 '21 at 17:30
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    @MarkViola Since it is not a positive term series, finding an upper bound wouldn't be sufficient to conclude. – CHAMSI Mar 15 '21 at 17:37
  • @CHAMSI The series on the right-hand side IS a series of positive terms. – Mark Viola Mar 15 '21 at 17:39
  • Use Dirichlet or Leibnitz test. – Koro Mar 15 '21 at 17:41
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    @CHAMSI that can easily be dealt with, at least intuitively. Group the terms of the series together in groups of two consecutive terms, starting off with $(1-1/2)+(1/3-1/4)+\cdots$ and each bracket is certainly positive. I don't know if that's rigorous or not though (although I can;t see why that shouldn't be sufficient). So the series certainly converges to value between $0$ and $0.75$. – A-Level Student Mar 15 '21 at 17:42
  • @A-LevelStudent Usually when a series has positive terms, the sequence of partial sums is increasing that is way finding an upper bound is sufficient to conclude. On the other hand, when a series is alternating, it is not monotone, and that argument does not stand by itself anymore, but I agree we can deal with it. – CHAMSI Mar 15 '21 at 18:00

2 Answers2

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The integral test cannot be used here, here is another way of proving the convergence.

Let $ n\geq 1 $, we have the following : \begin{aligned}\sum_{k=1}^{n}{\frac{\left(-1\right)^{k-1}}{k}}&=\int_{0}^{1}{\sum_{k=1}^{n}{\left(-x\right)^{k-1}}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\frac{1-\left(-1\right)^{n}x^{n}}{1+x}\,\mathrm{d}x}\\ \sum_{k=1}^{n}{\frac{\left(-1\right)^{k-1}}{k}}&=\ln{2}-\left(-1\right)^{n}\int_{0}^{1}{\frac{x^{n}}{1+x}\,\mathrm{d}x}\end{aligned}

Since : $ \left\vert\int_{0}^{1}{\frac{x^{n}}{1+x}\,\mathrm{d}x}\right\vert\leq\int_{0}^{1}{x^{n}\,\mathrm{d}x}=\frac{1}{n+1}\underset{n\to +\infty}{\longrightarrow}0$, then the sequence $ \left(\sum\limits_{k=1}^{n}{\frac{\left(-1\right)^{k-1}}{k}}\right)_{n\geq 1} $ converges, which means the series $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n-1}}{n}} $ converges. Also we have : $$ \sum_{n=1}^{+\infty}{\frac{\left(-1\right)^{n-1}}{n}}=\ln{2} $$

CHAMSI
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  • What is the star for in $\mathbb{N}^*$? – A-Level Student Mar 15 '21 at 17:26
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    @A-LevelStudent $ \mathbb{N}^{*}=\mathbb{N}\setminus\left\lbrace 0\right\rbrace$. It is a commonly used notation in French-speaking countries. – CHAMSI Mar 15 '21 at 17:27
  • @A-LevelStudent I assume not everyone will recognize it, so I'll change it to $ n\geq 1 $. – CHAMSI Mar 15 '21 at 17:32
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    Oh, I see. In other countries (mine included) the set $\mathbb{N}$ anyways only includes the positive (non-zero) integers. Alternatively, you could write $\mathbb{Z^+}$ for the positive (non-zero) integers. – A-Level Student Mar 15 '21 at 17:36
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Yes, it is an alternating series, and by the alternating series test, it does indeed converge!

You can verify it yourself, by substituitng $x=1$ into the Maclaurin series for $\ln( x+1)$, and that it converges to $\ln(2)$.

Ishraaq Parvez
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