Does this limit exist? 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?
 A: 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.
