# I Know that $\sum_{n=1}^\infty \frac{1}{n}$ Diverges, but what is $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ [duplicate]

We know that $\sum_{n=1}^\infty \frac{1}{n}$ diverges since it is a harmonic series. However, I was recently working on a homework problem where I was given to find if $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges. I found that it does converge, but I wanted to go further than just this. I was attempting to find what this sum was. I rechecked my work multiple times, and I know for sure that the sum does converge. How would I go about finding what the sum is for: $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$$

## marked as duplicate by Yagna Patel, Elliot G, drhab, Clement C., Patrick StevensDec 20 '15 at 19:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 2 Answers

Hint: Take the geometric sequence $$\frac 1{1-x}=\sum_{n=0}x^n,\quad\text{for } |x|<1.$$

Now, integrate both sides. You can integrate the right hand side term by term. You should see something familiar.

• But this only gives something legit under the restriction that $\lvert x\rvert <1$. As in the other answer, for the final step you need to invoke Abel's theorem, which is not a trivial step. – Clement C. Dec 20 '15 at 19:25
• Yes, that is true. – Tim Raczkowski Dec 20 '15 at 19:31

We have $$\log(1+x) = \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^n}n$$ for $\vert x \vert <1$ and $x=1$. Plug in $x=1$, to obtain the value of the expression.

• That requires a little bit more arguing. This is not clear that you can just "plug in" $x=1$, when the radius of convergence of the series is actually $1$... you need to call in some theorem for that. – Clement C. Dec 20 '15 at 19:23
• @ClementC. The interval of convergence is $(-1,1]$. And that is trivially obvious since the alternating series is monotonically decreasing. – Mark Viola Dec 20 '15 at 19:25
• All I am saying is that, no matter easy to show it, it is not "obvious": there is something to at least acknowledge. Failing to say there is something to argue may lead the OP (or the reader) to think one can plug in any $x$... – Clement C. Dec 20 '15 at 19:26
• @ClementC. What I have written is a true statement and in fact is true even when $x \in \mathbb{C}$ and lies inside and on the unit circle, except at $x=-1$. This follows trivially from the generalised alternating test. So you can indeed plug in $x=1$. No arguing is needed. – Leg Dec 20 '15 at 19:35
• @Leg the OP states "I was recently working on a homework problem where I was given to find if [the series] converges." Not sure if using consequences of properties of power series combined with the generalized alternating test counts as common knowledge then. Again, I do not claim the answer is incorrect. – Clement C. Dec 20 '15 at 19:57