I know by definition that a series $$\sum_{n=1}^\infty a_n$$ converges when the sequence of partial sums $S_N=a_1 + a_2 + .. + a_N$ converges to $S$, so $\lim_{N\rightarrow \infty} S_N=S$.

So in particular, I'm given the series $$\sum_{n=1}^\infty \frac{(-1)^n}{n},$$ which converges to $\ln 2$.

I'm kinda stuck on how to get started. So far I have,

$\forall \epsilon >0$, $\exists N>0$ s.t. if $n>N$ then $|a_n-0|<\epsilon$. Is this ok so far? How do I go from here?

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    $\begingroup$ "and I know it converges to 0." It is not true! The series converges to $\ln 2$ $\endgroup$ – Salech Rubenstein Dec 6 '13 at 4:20
  • $\begingroup$ It is true that the terms go to zero, but the partial sums do not go to zero. $\endgroup$ – vadim123 Dec 6 '13 at 4:20
  • $\begingroup$ It may help to split the sum into a sum of positive terms and one of negative terms, and show separately that each converges. $\endgroup$ – user99680 Dec 6 '13 at 4:20
  • $\begingroup$ @user99680 Bad luck, this only works for absolutely convergent series, which it is not (you would find that both diverge). You need the alternating series test, or use Mike's trick below. $\endgroup$ – Jean-Claude Arbaut Dec 6 '13 at 12:28

Your series does not converge to zero - the sequence it sums does, though. (You don't need to prove this, but your series sums to $-\text{ln}(2)$.)

Hint: Try grouping successive terms together. In particular, this becomes

$$\sum_{n=0}^\infty \left(-\frac 1 {2n+1} + \frac 1 {2n+2}\right)$$

See if you can work with that.

  • $\begingroup$ Ok, thanks! So now, I'll have two separate series of evens and odds, I think now I'm stuck on how to prove it using the sequence of partial sums (what I was stuck on before, but I had the series convergence incorrect.) $\endgroup$ – user12279 Dec 6 '13 at 12:44
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    $\begingroup$ Actually, you don't want two separate series. The harmonic series (sum over $1/n$) diverges. Instead, find a simpler way to write this... Hint: $1/n-1/(n+1)=1/(n^2+n)$ $\endgroup$ – user98602 Dec 6 '13 at 13:45

Now that you know the answer is $\text{ln}(2)$, perhaps you want to try some kind of estimate using Taylor's Theorem... Note that $\xi>0$ and so $1+\xi > 1$. $$|\sum_{k=1}^{n} \frac{(-x)^k}{k}-\text{ln}(1+x)| \leq \frac{\frac{n!}{(1+\xi)^n}\cdot x^{n+1}}{(n+1)!} < \frac{x^{n+1}}{n+1}$$

So in our case (it works in all possible values of $x$, since $x$ itself is fixed), we have $x=1$ and hence we have an upper bound of our error of $\frac{1}{n+1}$. Hence, as $n \rightarrow \infty$, $$\sum_{k=1}^{n} \frac{(-x)^k}{k}-\text{ln}(1+x) \rightarrow 0.$$

As for showing the sequence, $[S_{1},S_{2},...$, is convergent, just look at how $S_{1} < S_{3} < ... S_{2n+1}$ and $S_{2} > S_{4} > ... S_{2n}$ and combine to show that $S_{1} < S_{3} < ... S_{2n+1} < S_{2n} < S_{2n-2} < ... S_{4} < S_{2}$. Now, $\forall k,l>2n+1$, $|S_{k}-S_{l}| < S_{2n} - S_{2n+1} = \frac{1}{2n}$. So choose $n>\frac{1}{2\varepsilon}$ to complete the proof.

AFTERTHOUGHT: A subtle point is that I've proven Cauchy and not convergence here; however, in $\mathbb{R}$ they are equivalent so the proof still holds.

  • $\begingroup$ @user12279: does this help? $\endgroup$ – Chris K Dec 6 '13 at 5:11
  • $\begingroup$ sorry, I'm not familiar with Taylor's Theorem. I was trying to I guess prove it from the definition. $\endgroup$ – user12279 Dec 6 '13 at 12:43
  • $\begingroup$ @user12279: Note that Taylor's Theorem is used to prove that the series converges to $ln(2)$, but that is a much stronger statement than what you are probably looking for; that is the series converges. For this see the paragraph afterwards. I give you a proof that works with your definition. $\endgroup$ – Chris K Dec 6 '13 at 16:28

The following lemma (known in some circles as Leibniz's test) trivially solves the problem.

Let $(a_n)_{n=1}^{\infty}$ be a monotonic sequence of real numbers converging to $0$. Then the following series $$ \sum_{n=1}^{\infty} (-1)^n a_n $$ converges.

I leave out the proof, but it can be found on Wikipedia.


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