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I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges, but what about the alternating harmonic series

$$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}$$

Does it converge? If so, what is its sum?

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    $\begingroup$ $$\ln(1-x)=-\sum_{n=1}^\infty\frac{x^n}n\quad\iff\quad\sum_{n=1}^\infty\frac{(-1)^{n+1}}n=-\sum_{n=1}^\infty\frac{(-1)^n}n=\ln(1-[-1])=\ln2$$ $\endgroup$ – Lucian Nov 19 '13 at 7:39
  • $\begingroup$ I know a non-calculus way, but you have to assume $e^x\ge x+1$. Would you like me to share it with you? $\endgroup$ – Akiva Weinberger Mar 24 '15 at 17:37
  • $\begingroup$ @AkivaWeinberger If not the OP, I am most certainly interested. $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 22:38
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Complementary to Mau's answer:

Call a series $a_n$ absolutely convergent if $\sum|a_n|$ converges. If $a_n$ converges but is not absolutely convergent we call $a_n$ conditionally convergent The Riemann series theorem states that any conditionally convergent series can be reordered to converge to any real number.

Morally this is because both the positive and negative parts of your series diverge but the divergences cancel each other out, one or other's canceling the other can be staggered by adding on, say, the negative bits every third term in stead of every other term. This means that in the race for the two divergences to cancel each other out, we give the positive bit something of a head-start and will get a larger positive outcome. Notice how, even in this rearranged version of the series, every term will still come up exactly once.

It is also worth noting, on the Wikipedia link Mau provided, that the convergence to $\ln 2$ of your series is at the edge of the radius of convergence for the series expansion of $\ln(1-x)$- this is a fairly typical occurrence: at the boundary of a domain of convergence of a Taylor series, the series is only just converging- which is why you see this conditional convergence type behavior.

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    $\begingroup$ This doesn't seem to answer the question. $\endgroup$ – Teepeemm Apr 3 '17 at 19:13
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There are actually two "more direct" proofs of the fact that this limit is $\ln (2)$.

First Proof Using the well knows (typical induction problem) equality:

$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n} \,.$$

The right side is $\frac{1}{n} \left[ \frac{1}{1+\frac{1}{n}}+ \frac{1}{1+\frac{2}{n}}+..+\frac{1}{1+\frac{n}{n}} \right]$ which is the standard Riemann sum associated to $\int_0^1 \frac{1}{1+x} dx \,.$

Second Proof Using $\lim_n \frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln (n) =\gamma$.

Then

$$\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2n-1}-\frac{1}{2n}= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n} \right]-2 \left[\frac{1}{2}+\frac{1}{4}...+\frac{1}{2n} \right] $$

$$= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2n-1}+\frac{1}{2n} \right]-\ln(2n) - \left[\frac{1}{1}+\frac{1}{2}...+\frac{1}{n} \right]+\ln(n) + \ln 2 \,.$$

Taking the limit we get $\gamma-\gamma+\ln(2)$.

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  • $\begingroup$ First equation, right-hand side: Should the first fraction be $\frac{1}{n+1}$? $\endgroup$ – Mike Spivey Jun 2 '11 at 5:09
  • $\begingroup$ I don't consider the second a proof because it is totally not apparent why that limit should exist, and it is no easier. $\endgroup$ – user21820 Apr 3 '15 at 11:35
  • $\begingroup$ @user21820 The existence of the limit $\lim_n \frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln (n)$ is well known and an easy consequence of the estimates of the integral test. $\endgroup$ – N. S. Apr 3 '15 at 15:05
  • $\begingroup$ @N.S.: Yes indeed but so is $\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\cdots$ well known and provable without even using the existence of $\gamma$ that you used. My point is that what you left unproven is as hard or even harder than the original problem. $\endgroup$ – user21820 Apr 4 '15 at 5:07
  • $\begingroup$ @user21820 Harder is relative. This limit together with the standard $\lim_n (1+\frac{1}{n})^n$ were the two first non-trivial limits I actually learned in highschool... And I don't think the proof is that hard : $$a_n = \frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln (n) \,;\, b_n = \frac{1}{1}+\frac{1}{2}+...+\frac{1}{n}-\ln (n+1)$$ then $b_n \leq a_n$ is trivial and $a_n$ decreasing, $b_n$ increasing are immediate... $\endgroup$ – N. S. Apr 4 '15 at 15:41
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In this answer, I used only Bernoulli's inequality to show that $$ \left(\frac{2n+1}{n+1}\right)^\frac{n}{n+1} \le\left(1+\frac1n\right)^{n\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)} \le\frac{2n+1}{n+1}\tag{1} $$ The squeeze theorem and $(1)$, show that $$ \exp\left[\lim\limits_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\right]=2\tag{2} $$ That is, $$ \begin{align} \lim_{n\to\infty}\left(1-\frac12+\frac13-\frac14+\dots-\frac1{2n}\right) &=\lim_{n\to\infty}\left(\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}\right)\\[6pt] &=\log(2)\tag{3} \end{align} $$

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  • $\begingroup$ Is it the change in form of the series, across the = in the line above (3), that is dependent on the ordering of the series? $\endgroup$ – Isaac Apr 16 '13 at 16:14
  • $\begingroup$ @Isaac: The stuff inside the parentheses is equal. There is no problem about the ordering when summing a finite number of terms. $\endgroup$ – robjohn Apr 16 '13 at 16:21
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    $\begingroup$ This is really very unusual and new for me. +1. $\endgroup$ – Paramanand Singh Mar 29 '16 at 4:19
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    $\begingroup$ @xFioraMstr18: Both left and right side of $(1)$ tend to $2$; therefore, the middle part of $(1)$ tends to $2$ by the Squeeze Theorem. Now we know that $\left(1+\frac1n\right)^n$ can be made as close to $e$ as we wish by setting $n$ large enough. Therefore, the middle part of $(1)$ can be made as close to $(2)$ as we wish. $\endgroup$ – robjohn Jun 29 '18 at 12:33
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    $\begingroup$ If that is too many things "close to ..." for you, we can take the log of $(1)$ divided by $n\log\left(1+\frac1n\right)$: $$ \frac{\overbrace{\ \ \ \ \frac{n}{n+1}\vphantom{\frac21}\ \ \ \ }^{\to1}\overbrace{\log\left(\frac{2n+1}{n+1}\right)}^{\to\log(2)}}{\underbrace{n\log\left(1+\frac1n\right)}_{\to1}} \le\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n} \le\frac{\overbrace{\log\left(\frac{2n+1}{n+1}\right)}^{\to\log(2)}}{\underbrace{n\log\left(1+\frac1n\right)}_{\to1}} $$ Thus, the middle part tends to $\log(2)$ by the Squeeze Theorem. $\endgroup$ – robjohn Jun 29 '18 at 12:33
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it is not absolutely convergent (that is, if you are allowed to reorder terms you may end up with whatever number you fancy).

If you consider the associated series formed by summing the terms from 1 to n of the original one, that is you fix the order of summation of the original series, that series (which is not the original one...) converges to $\ln(2)$ See Wikipedia.

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  • $\begingroup$ But rearranging terms on pretty much anything that goes out infinitely allows any possible answer, hence changing what you are summing. $\endgroup$ – Simply Beautiful Art Jan 7 '16 at 22:55
  • $\begingroup$ this is why I wrote "If you consider the associated series [...]" and I computed that sum. $\endgroup$ – mau Jan 8 '16 at 12:14
  • $\begingroup$ I just want future readers to understand your answer better. $\endgroup$ – Simply Beautiful Art Jan 8 '16 at 22:00
  • $\begingroup$ and indeed my first sentence says "(that is, if you are allowed to reorder terms you may end up with whatever number you fancy)." $\endgroup$ – mau Jan 9 '16 at 9:53
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Let's say you have a sequence of nonnegative numbers $a_1 \geq a_2 \geq \dots$ tending to zero. Then it is a theorem that the alternating sum $\sum (-1)^i a_i$ converges (not necessarily absolutely, of course). This in particular applies to your series.

Incidentally, if you're curious why it converges to $\log(2)$ (which seems somewhat random), it's because of the Taylor series of $\log(1+x)$ while letting $x \to 1$.

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    $\begingroup$ To add to Akhil's answer, one needs to invoke Abel's theorem (en.wikipedia.org/wiki/Abel%27s_theorem), since 1 is at the border of the interval of convergence. This is a delicate test that ensures that the numerical series converges to the number the power series predicts. $\endgroup$ – Andrés E. Caicedo Jun 2 '11 at 5:12
  • $\begingroup$ @Andres: Thanks for fixing my grammar and for the comment! $\endgroup$ – Akhil Mathew Jun 6 '11 at 13:22
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$\sum_{k=1}^{n} ( \frac{1}{2k-1}-\frac{1}{2k} ) = \sum_{k=1}^{n} ( \frac{1}{2k-1}+\frac{1}{2k} ) - 2 \sum_{k=1}^{n} \frac{1}{2k} = \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k} = \sum_{k=n+1}^{2n} \frac{1}{k}$.

$\ln(2) \overset{n\to\infty}{\leftarrow} \ln(2) + \ln(\frac{2n+1}{2n+2}) = \ln(2n+1)-\ln(n+1)$

$= \int_{n+1}^{2n+1} \frac{1}{x}\ dx \le \sum_{k=n+1}^{2n} \frac{1}{k} \le \int_{n}^{2n} \frac{1}{x}\ dx$

$= \ln(2n)-\ln(n) = \ln(2)$.

So by squeeze theorem we are done.

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Here is another proof, based on the formula

$$\frac{1}{1+x}=\frac{(-x)^n}{1+x}+\sum_{k=0}^n(-x)^k$$

Integrating both sides over $[0,t]$ gives

$$\ln(1+t)=\int_0^t\frac{(-1)^nx^n}{1+x}\,dx+\sum_{k=1}^n\frac{(-t)^{k+1}}{k}$$

Setting $t=1$ shows that the partial sums $s_n$ of the alternating harmonic series are given by

$$s_n=\ln2-(-1)^n\int_0^1\frac{x^n}{1+x}\,dx$$

But on $[0,1]$, we have $0\leq x^n(1+x)^{-1}\leq x^{n-1}$, so

$$0\leq\int_0^1\frac{x^n}{1+x}\,dx\leq\int_0^1x^{n-1}\,dx=\frac{1}{n}$$

Hence $$\ln2-\frac{1}{n}\leq s_n\leq\ln2+\frac{1}{n}$$ So $s_n\to\ln 2$ as $n\to\infty$.

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  • $\begingroup$ $\sum_{k=0}^n(-x)^k= \frac{1-(-x)^n}{1+x}$. You're missing some $(-1)^n$ at several places. Also, $s_n=\ln2-\int_0^1\frac{x^n}{1+x}\,dx$ cannot be right because $ s_n-\ln2$ alternates in sign. $\endgroup$ – Gabriel Romon Sep 12 '18 at 8:18
  • $\begingroup$ @GabrielRomon: Thanks for pointing out the problems; I wrote this up too quickly, probably on my phone. Fixed now. $\endgroup$ – symplectomorphic Sep 12 '18 at 20:29
  • $\begingroup$ Of course, this begs the question what is the inspiration for the original formula? Easily verified by induction, however it does go against a student's habit of partial decomposition in that you're making the numerator more complicated to get more polynomial out front. I suspect it's based upon $1/(1-x)=1+x+x^2+x^3+\cdots$ and scaled to cut this off at some point. Is there further explanation for the inspiration? $\endgroup$ – Robert Wolfe Dec 21 '18 at 6:22
  • $\begingroup$ (1) “begs the question” isn’t the phrase you want; that means the argument is circular. You mean “raises the question. (2) No, the formula isn’t based on an infinite series; just do the polynomial long division. $\endgroup$ – symplectomorphic Dec 21 '18 at 14:50
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    $\begingroup$ @NewBornMATH: just like Robert Wolfe above, you are not reading my argument correctly; I don’t use the infinite geometric series for $1/(1+x)$. All I do is integrate a formula derived from (finite) polynomial division, and then estimate the integral that results. $\endgroup$ – symplectomorphic Jan 23 at 18:54
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I want to use the infinite series expansion of the integral of a function to compute the sum. If you repeat the process of integration by parts over and over again: $$\int f(x) dx = xf(x)-\int xf'(x)dx = xf(x) - \frac{x^2}{2}f'(x)+\int \frac{x^2}{2}f''(x)dx$$ $$\Rightarrow \int f(x) dx = xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\int \frac{x^3}{6}f''(x)dx$$ $$\Rightarrow \int f(x) dx = xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\frac{x^4}{24}f'''(x)+ \int \frac{x^4}{24}f'''(x)dx$$ Continuing this pattern we can prove using Mathematical Induction (for all $n\ \epsilon\ \mathbb{Z}^+$)that: $$\int f(x)dx = \sum_{k=0}^{n}\left[\frac{(-1)^kf^{(k)}(x)x^{k+1}}{(k+1)!}\right]+\frac{(-1)^n}{n!}\int x^n f^{(n)}(x)dx \rightarrow (1)$$ Limiting both sides of equation (1) as $n \rightarrow \infty$ we can write the infinite series expansion: $$\int f(x)dx = \sum_{k=0}^{\infty}\left[\frac{(-1)^kf^{(k)}(x)x^{k+1}}{(k+1)!}\right] + C=C+xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-+...\rightarrow (2)$$ We can certainly remove the integration constant $C$ from both sides by taking the definite integral: $$\int_{0}^{x} f(x)dx =xf(x) - \frac{x^2}{2}f'(x) + \frac{x^3}{6}f''(x)-\frac{x^4}{24}f'''(x)+-...\rightarrow (3)$$ We can prove the Taylor series expansion for functions using this idea. Now let $f(x)=\frac{1}{1+x}$ and we get: $$\ln(1+x)=\frac{1}{1+\frac{1}{x}}+\frac{1/2}{(1+\frac{1}{x})^2}+\frac{1/3}{(1+\frac{1}{x})^3}+\frac{1/4}{(1+\frac{1}{x})^4}+\frac{1/5}{(1+\frac{1}{x})^5}+...\rightarrow (4)$$ $$y=\frac{x}{x+1} \Rightarrow \ln (\frac{1}{1-y})=y+\frac{y^2}{2}+\frac{y^3}{3}+\frac{y^4}{4}+\frac{y^5}{5}+...\rightarrow (5)$$ By letting $y=-1$ and multiplying both sides of equation $(5)$ by $-1$ we get the sum we desired: $$\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5}-\frac{1}{6}+-...$$ P.S. You can also use $(4)$ to get another interesting expansion for $\ln 2$ by putting $x=1$: $$\ln 2 = \frac{1}{1 \cdot 2^1}+\frac{1}{2 \cdot 2^2}+\frac{1}{3 \cdot 2^3}+\frac{1}{4 \cdot 2^4}+\frac{1}{5 \cdot 2^5}+...$$

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A proof without words by Matt Hudleson proof

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