A pair of limits involving sums Find the following pair of limits:
$$\lim_{r \to \infty} 2 \lfloor r \rfloor \sum_{n=2\lfloor r \rfloor+1}^\infty \frac{(-1)^{n+1}}{n}$$
$$\lim_{r \to \infty} (2 \lfloor r \rfloor +1) \sum_{n=2 \lfloor r \rfloor+2}^\infty \frac{(-1)^{n+1}}{n}$$
Plugging in large values of $r$, I would guess that the first limit tends to $1/2$ and that the second limit tends to $-1/2$.
The floors are intended to require that $r$ be an integer. The inside of the first limit is just the following function for even $r$ and the inside of the second is the function for odd $r$.
$$r \sum_{n=r+1}^\infty  \frac{(-1)^{n+1}}{n}$$
Normally for a problem like this, I'd try L'Hopital's rule, however I don't know how that would work because the obvious way would require taking the derivative
$$\frac{d}{dr} \sum_{n=2\lfloor r \rfloor+1}^\infty \frac{(-1)^{n+1}}{n}$$
which doesn't seem to make very much sense.
Any hints or solutions on how to evaluate these limits?
 A: For $r\in \Bbb N$  we have $$\sum_{n=2r+1}^{\infty}(-1)^{n+1}/n=$$ $$=(\frac {1}{2r+1}-\frac {1}{2r+2})+(\frac {1}{2r+3}-\frac {1}{2r+4})+(\frac {1}{2r+5}-\frac {1}{2r+6})+...=$$ $$=\frac  {1}{(2r+1)(2r+2)}+\frac {1}{(2r+3)(2r+4)}+\frac {1}{(2r+5)(2r+6)}+...$$
Now for $n>1$ we have $$\frac {1}{4}\int_{n+1}^{n+2}(1/x^2)dx<\frac {1}{4(n+1)^2}<$$ $$<\frac {1}{(2n+1)(2n+2)}<$$ $$<\frac {1}{(2n+1)^2}<\frac {1}{4n^2}<$$ $$<\frac {1}{4}\int_{n-1}^n(1/x^2)dx.$$
Summing this for $n$ from $r$ to $\infty$ ( with $1<r\in \Bbb N$ ) we have  $$\frac {1}{4}\int_{r+1}^{\infty}(1/x^2)dx<\sum_{n=2r+1}^{\infty}(-1)^{n+1}/n<\frac {1}{4}\int_{r-1}^{\infty}(1/x^2)dx.$$
The far left and far right sides in the line above are $\frac {1}{4(r+1)}$ and $\frac {1}{4(r-1)}.$
The second limit can be done similarly but it can also be obtained from the first limit, for if $F(s)=\sum_{n=s}^{\infty}(-1)^{n+1}/n$ then $$(2r+1)F(2r+2)=\frac {2r+1}{2r}\cdot 2r\left(F(2r+1)-\frac {1}{2r+1}\right)= \frac {2r+1}{2r}\cdot 2rF(2r+1)-1.$$ 
