How to evaluate the series: $$ 2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$

According to Mathematica, this converges to $ (\log 2)^2 $.

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    $\begingroup$ What is the summand when $n=0$? Is it just $-1$? Or perhaps the summation runs from $n=1$ to $n = \infty$ instead? $\endgroup$ Feb 2 '13 at 18:47
  • $\begingroup$ For large $n$ the absolute value of adjacent terms is $ n\log{(n+1)}/((n+1) \log{n}) \sim (n \log{n}+1)/(n \log{n} +n)$ which does go to zero. $\endgroup$
    – Ron Gordon
    Feb 2 '13 at 18:48
  • $\begingroup$ For $|a_{n+1}|\leq |a_n|$, write the inequality down, multiply by $n+2$, then use $\frac{n+2}{n+1}=1+\frac{1}{n+1}$. It works. $\endgroup$
    – Julien
    Feb 2 '13 at 18:50
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    $\begingroup$ @experimentX you answered your own question... your sum is exactly equal to the product $$\left(\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{n+1}\right)^2 = (-\log 2)^2 = (\log 2)^2.$$ $\endgroup$ Feb 2 '13 at 19:37
  • 1
    $\begingroup$ Hint: If $f(x)=(\log x)^2$, then $f^{(n+1)}(1)=2\,(-1)^{n+1}(n!)(1+\frac{1}{2}+\cdots+\frac{1}{n})$. See related question $\endgroup$
    – Michael E2
    Feb 2 '13 at 19:56

Recall that, formally,

$$ \left(\sum_{n=1}^{\infty} a_n\right)\left(\sum_{n=1}^{\infty} b_n\right) = \sum_{n=1}^{\infty} c_{n+1},$$


$$ c_n = \sum_{k=1}^{n-1} a_k b_{n-k}. $$

If the series $\sum c_{n+1}$ converges, then the above equality is actually true. You seem to know how to show this, so I'll just demonstrate the formal aspect of the problem.

Let $a_n = b_n = \frac{(-1)^{n}}{n}$. Then

$$ a_k b_{n-k} = \frac{(-1)^n}{k(n-k)} = \frac{(-1)^n}{n}\left(\frac{1}{k}+\frac{1}{n-k}\right), $$

so that

$$ \begin{align*} c_n &= \frac{(-1)^n}{n} \sum_{k=1}^{n-1} \left(\frac{1}{k}+\frac{1}{n-k}\right) \\ &= 2\frac{(-1)^n}{n} \sum_{k=1}^{n-1} \frac{1}{k}. \end{align*} $$

We therefore have

$$ 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n+1} \sum_{k=1}^{n} \frac{1}{k} = \left(\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)^2 = (-\log 2)^2 = (\log 2)^2. $$

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    $\begingroup$ wow!! i only knew one had to absolutely convergent. $\endgroup$ Feb 2 '13 at 20:15
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    $\begingroup$ @experimentX, You may be interested in these notes (the bottom of p.4), which proves that result using Abel's theorem. $\endgroup$ Feb 2 '13 at 20:29
  • $\begingroup$ For searching purposes: what Antonio has used here is the fact that the Cauchy product of the generating functions of two independent sequences is the generating function of the convolution of the two sequences. (+1, of course.) $\endgroup$ Apr 3 '13 at 7:36
  • $\begingroup$ For an updated reference, see Theorem 14.17 in Pete L. Clark's Honors Calculus notes. $\endgroup$ May 19 '15 at 19:37

Use generating functions:

Consider $$-\log(1-x) = \sum_{n=1}^\infty \frac{x^n}{n}.$$ Dividing by $1-x$, we get $$-\frac{\log(1-x)}{1-x} = \sum_{n=1}^\infty \left(\sum_{k=1}^n \frac{1}{k}\right)x^n.$$ Integrating this and multiplying everything by $2$ gives $$\left[\log(1-x)\right]^2 = 2\sum_{n=1}^\infty \left(\sum_{k=1}^n \frac{1}{k}\right)\frac{x^{n+1}}{n+1} + C,$$ where $C$ is some constant. But we can get rid of $C$ by plugging $x=0$ into both sides, which gives $C=0$: $$\left[\log(1-x)\right]^2 = 2\sum_{n=1}^\infty \left(\sum_{k=1}^n \frac{1}{k}\right)\frac{x^{n+1}}{n+1}.$$ From here, we'd like to simply plug in $x=-1$ and say our answer is $(\log{2})^2$, but we have to first check to make sure the power series on the right actually converges there. To do this, set $H_n=1+\frac{1}{2}+\cdots + \frac{1}{n}$ (the "$H$" is for "harmonic", since $H_n$ is the $n$th harmonic number). Let's see when the inequality $$ \frac{(n+1)H_{n+1}}{(n+2)H_n}<1$$ holds. Rearranging terms, and using the fact that $H_{n+1}=H_n+\frac{1}{n+1}$, it follows that the above inequality holds exactly when $H_n>1$. But a quick glance at the definition of $H_n$ shows that this is always true! Therefore, the terms of our series decrease in absolute value. Since they also converge to zero (they're all less than $1/(n+1)$, which converges to zero), the entire series converges by the alternating series test.

  • $\begingroup$ wow!! this is also very nice!! $\endgroup$ Feb 2 '13 at 20:34
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    $\begingroup$ You are implicitly using Abel's theorem: en.wikipedia.org/wiki/Abel%27s_theorem $\endgroup$
    – Aryabhata
    Feb 2 '13 at 20:39
  • $\begingroup$ Nice answer, easy way to follow! (+1) $\endgroup$ Feb 2 '13 at 20:42
  • $\begingroup$ The line with the 1st inequality sign in this answer is wrong.We are reducing the numerator by 1 so the RHS should be lesser than the LHS. $\endgroup$ Dec 29 '13 at 1:37
  • $\begingroup$ @Jack'swastedlife I'm surprised I got 13 upvotes before anyone noticed that! I've (hopefully) fixed it, and in the process actually ended up in my opinion simplifying the convergence argument. Thanks for pointing it out. $\endgroup$ Dec 29 '13 at 23:40

This is a special case of a more general result derived here.

$$S = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n+1} \sum_{k=1}^n \dfrac1k$$ Recall that $\dfrac1k = \displaystyle \int_0^1 x^{k-1} dx$ and $\dfrac1{n+1} = \displaystyle \int_0^1 y^n dy$.

Now use the following fact. $$\sum_{k=0}^{\infty} \int_0^1 (-z)^k dz = \lim_{n \to \infty} \int_0^1 \dfrac{1 - (-z)^n}{1+z} dz$$ The sequence of functions $f_n(z) = \dfrac{1 - (-z)^n}{1+z}$ is dominated by the function $g(z) = \dfrac2{1+z}$ in the interval $[0,1]$, which is integrable. Hence, we can swap the limit and the integral to get that $$\lim_{n \to \infty} \int_0^1 \dfrac{1 - (-z)^n}{1+z} dz = \int_0^1 \dfrac{dz}{1+z}$$

Hence, $$S = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 y^n dy \left(\sum_{k=1}^n \int_0^1 x^{k-1} dx \right) = \sum_{n=1}^{\infty} (-1)^{n+1} \int_0^1 y^n dy \left(\int_0^1 \dfrac{1-x^n}{1-x} dx \right)$$ Hence, $$S = \int_0^1 \int_0^1 \dfrac{\dfrac{y}{1+y} - \dfrac{xy}{1+xy}}{1-x} dy dx = \int_0^1 \int_0^1 \dfrac{y+xy^2-xy-xy^2}{(1+y)(1+xy)(1-x)} dx dy\\ =\int_0^1 \int_0^1 \dfrac{y}{(1+y)(1+xy)} dx dy = \int_0^1 \dfrac{\log(1+y)}{1+y} dy = \left. \dfrac{\log^2(1+y)}2 \right \vert_0^1 = \dfrac{\log^2(2)}2$$ The sum you are interested in is $2S$ and hence the answer is $\log^2(2)$.

  • $\begingroup$ Beautiful answer, but how do you justify swapping the infinite sum and integrals? $\endgroup$
    – PeterM
    Feb 2 '13 at 21:49
  • $\begingroup$ @PeterM I have clarified this by updating the answer. Let me know if I have missed something out. $\endgroup$
    – user17762
    Feb 2 '13 at 22:08
  • $\begingroup$ sweet indeed - there's a small typo $\sum_{k=0}^{\infty} \int_0^1 (-z)^k dy$ should be $\sum_{k=0}^{\infty} \int_0^1 (-z)^k dz$ $\endgroup$
    – nikola
    Feb 6 '13 at 17:54
  • $\begingroup$ @nikola Thanks. Corrected. $\endgroup$
    – user17762
    Feb 6 '13 at 17:58
  • $\begingroup$ @Marvis Perfect. Thank you! $\endgroup$
    – PeterM
    Feb 6 '13 at 23:13

In this answer, it is shown that $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n=\frac12\zeta(2)-\frac12\log(2)^2 $$ This sum is $$ \begin{align} 2\sum_{n=1}^\infty\frac{(-1)^n}{n}H_{n-1} &=2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\left(\frac1n-H_n\right)\\ &=2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}-2\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}H_n\\ &=\zeta(2)-\left(\zeta(2)-\log(2)^2\right)\\[6pt] &=\log(2)^2 \end{align} $$


Here is another approach that I just noticed $$ \begin{align} 2\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n+1}\sum_{k=1}^n\frac1k &=2\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{(-1)^{n+1}}{k(n+1)}\tag{1}\\ &=2\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^{n+k}}{k(n+k)}\tag{2}\\ &=2\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+k}\frac1n\left(\frac1k-\frac1{n+k}\right)\tag{3}\\ &=\sum_{k=1}^\infty\sum_{n=1}^\infty(-1)^{n+k}\frac1n\frac1k\tag{4}\\[6pt] &=\log(2)^2\tag{5} \end{align} $$ Explanation:
$(1)$: change the order of summation
$(2)$: substitute $n\mapsto n+k-1$
$(3)$: partial fractions
$(4)$: swap $n$ and $k$ in $(2)$ add to $(3)$ and divide by $2$
$(5)$: $\sum\limits_{n=1}^\infty\frac{(-1)^{n-1}}{n}=\log(2)$

  • $\begingroup$ Beautiful, step (4) is amazing $\endgroup$ Nov 25 '18 at 20:46

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &2\sum_{n = 1}^{\infty}{\pars{-1}^{n + 1} \over n + 1} \pars{1 + {1 \over 2} + \cdots + {1 \over n}} = -2\sum_{n = 1}^{\infty}\pars{-1}^{n}\, H_{n}\int_{0}^{1}x^{n}\,\dd x = -2\int_{0}^{1}\sum_{n = 1}^{\infty}H_{n}\pars{-x}^{n}\,\dd x \\[5mm] = &\ -2\int_{0}^{1}\braces{-\,{\ln\pars{1 - \bracks{-x}} \over 1 - \pars{-x}}}\,\dd x = 2\int_{0}^{1}{\ln\pars{1 +x} \over 1 + x}\,\dd x = \left.\ln^{2}\pars{1 + x}\,\right\vert_{\ x\ =\ 0}^{\ x\ =\ 1} = \bbx{\ds{\ln^{2}\pars{2}}} \end{align}


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