Estimating the tail of alternating harmonic sum I am trying to solve a problem from an old qualification exam and want to prove that $$\lim_{n\to\infty} n \sum_{i=2n}^\infty \frac{(-1)^i}{i} = \frac{1}{4}.$$
I know that the limit exists because the sum is bounded by $\frac{1}{2n}$ in magnitude, so multiplying by $n$ and taking the limit gives $\frac{1}{2}$ as an upper bound. I was hoping to get some kind of integral approximation for the sum, but because of the alternating signs, I'm not sure how to do that. 
 A: Here's an elementary approach:
$$\sum_{k=2n}^\infty\frac{(-1)^k}k=\sum_{k=n}^\infty\left [\frac1{2k}-\frac1{2k+1}\right]$$
By the integral test,
$$\int_n^\infty\left[\frac1{2x}-\frac1{2x+1}\right]~\mathrm dx<\sum_{k=n}^\infty\left [\frac1{2k}-\frac1{2k+1}\right]<\int_{n-1}^\infty\left[\frac1{2x}-\frac1{2x+1}\right]~\mathrm dx$$
These limits may easily be evaluated using the squeeze theorem and L'Hospital's rule, noting that
$$\lim_{n\to\infty}n\int_n^\infty f(x)~\mathrm dx=\lim_{n\to\infty}\frac{\displaystyle\int_n^\infty f(x)~\mathrm dx}{1/n}=\lim_{n\to\infty}\frac{f(n)}{1/n^2}$$
A: $$n \sum_{i=2n}^\infty \dfrac{(-1)^i}{i} = \frac{n}{2} \left(\Psi\left(n+\frac12\right) - \Psi(n)\right)$$ 
Now $$\eqalign{\Psi(n) &= \ln(n) - \frac{1}{2n} + O(1/n^2)\cr
\Psi\left(n+\frac12\right) &= \ln(n+1/2) - \frac{1}{2n+1} + O(1/n^2)\cr
&= \ln(n) + O(1/n^2)  }$$
so 
$$  \frac{n}{2} \left(\Psi\left(n+\frac12\right) - \Psi(n)\right) = \frac{1}{4} + O(1/n)$$
A: You may like this method. By the Stolz theorem,
\begin{eqnarray}
\lim_{n\to\infty} n \sum_{i=2n}^\infty \frac{(-1)^i}{i} &=&
\lim_{n\to\infty} \frac{\sum_{i=2n}^\infty \frac{(-1)^i}{i}}{\frac{1}{n}}\\
&=&\lim_{n\to\infty} \frac{\sum_{i=2(n+1)}^\infty \frac{(-1)^i}{i}-\sum_{i=2n}^\infty \frac{(-1)^i}{i}}{\frac1{n+1}-\frac1{n}}\\
&=&\lim_{n\to\infty}\frac{-\frac{1}{2n}+\frac{1}{2n+1}}{-\frac{1}{n(n+1)}}\\
&=& \frac{1}{4}
\end{eqnarray}
