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Does this limit exist?

$$ \lim_{n \rightarrow \infty}\sum_{(i=1)}^n (-1)^{i-1} \frac{1}{i} $$

Also, is there a name for this sum?

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Yes, by the alternating series test. Are you familiar with the Taylor series of the natural logarithm? – Qiaochu Yuan Sep 11 '11 at 2:06
It's sometimes referred to as the "alternating harmonic series". – J. M. Sep 11 '11 at 2:10
I think it equals to $\log 2$ – gaurav Sep 11 '11 at 2:31
up vote 8 down vote accepted

This series is sometimes referred to as the alternating harmonic series. (An alternating series is just a series with alternating nonnegative and nonpositive terms.)

The convergence of this series follows directly from the alternating series test or the Leibniz test, of which this series gives an ideal textbook example.

Alternating series test. Suppose that the sequence $\lbrace a_n \rbrace_{n \geq 1}$ is monotonically decreasing and converges to $0$. Then the alternating series $\sum \limits_{n=1}^{\infty} (-1)^{n-1} a_n$ converges.

On the other hand, it is not absolutely convergent, since the harmonic series $\sum \frac{1}{n}$ famously diverges again and again. (See also this question: 255.)

It turns out that we also know the exact sum of the series. The logarithmic function has the Taylor expansion: $$ \ln (1+x) = \sum\limits_{n=1}^{\infty} (-1)^{n-1} \frac{x^n}{n} \ \ \ \ (-1 \lt x \lt 1), $$ with a radius of convergence $1$. Since $1$ is the end-point of the interval of convergence, it is not immediately clear that one can plug in $1$ in the above series. However, Abel's theorem for power series, in fact, guarantees that $$ \lim_{x \to 1-} \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}. $$ Thus the series evaluates to $\ln 2 \approx 0.693\ldots$.

This answer is mostly an elaboration of Qiaochu Yuan and J.M.'s comments. Thanks to Robert Israel for pointing out the need for Abel's theorem.

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Since $1$ is an endpoint of the interval of convergence, it's not so obvious that the sum of your series is $\ln 2$. You need Abel's theorem. See e.g. – Robert Israel Sep 11 '11 at 6:12
@Robert You are correct. But I was not trying to derive the domain of convergence. I was more like, well, stating it from the know-it-all wikipedia. :-) I will update my answer to be more complete. – Srivatsan Sep 11 '11 at 6:16

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