Determine $N$ of series $\sum\limits^N_{n=1}(-1)^n\frac{1}{2n+1}$ so that it differs from the actual sum by less than $0.001$ 
Determine $N$ of series $\sum^N_\limits{n=1}(-1)^n\frac{1}{2n+1}$ so that it differs from $\sum\limits^\infty_{n=1}(-1)^n\frac{1}{2n+1}$ by less than $\frac1{1000}$

How do I do this problem?
I know that the remainder is equal to the sum to infinity - the sum to N should be less than 0.001 given by the formula
$$|R_n|=|S-S_n| \le b_{n+1}$$
$$\int^\infty_1\frac1{2x+1}-\int_N^\infty\frac1{2x+1} \le 0.001$$
$$\infty - \frac12\left[\ln(2n+1)\right]^\infty_N \le 0.001$$
Which doesn't really make any sense... at least to me
 A: Let
$$
R_N = \sum_{n\ge N+1} (-)^n{1\over 2n+1}
$$
observe that 
$$
|R_N| < {1\over 2N+1}
$$
then we would like
$$
\eqalign{
& R_N < {1\over 1000} \cr
\rightarrow\  & {1\over 2N+1} < {1\over 1000} \cr
\rightarrow\  & 2N+1 > 1000 \cr
\rightarrow\  & N \ge 500 \cr
}
$$
A: $$
\begin{align}
\sum_{k=2n}^\infty(-1)^k\frac1{2k+1}
&=\sum_{k=n}^\infty\left(\frac1{4k+1}-\frac1{4k+3}\right)\\
&=\sum_{k=n}^\infty\frac2{(4k+1)(4k+3)}\\
&\lt\sum_{k=n}^\infty\frac2{4k(4k+4)}\\
&=\frac12\sum_{k=n}^\infty\left(\frac1{4k}-\frac1{4k+4}\right)\\
&=\frac1{8n}
\end{align}
$$

Thus, at $n=125$, the error is less than $\frac1{1000}$.


$$
\begin{align}
\sum_{k=2n-1}^\infty(-1)^{k-1}\frac1{2k+1}
&=\sum_{k=n}^\infty\left(\frac1{4k-1}-\frac1{4k+1}\right)\\
&=\sum_{k=n}^\infty\frac2{(4k-1)(4k+1)}\\
&\gt\sum_{k=n}^\infty\frac2{(4k-1)(4k+3)}\\
&=\frac12\sum_{k=n}^\infty\left(\frac1{4k-1}-\frac1{4k+3}\right)\\
&=\frac1{8n-2}
\end{align}
$$
Thus, at $n=125$, the error is greater than $\frac1{1000}$.

Thus, the first partial sum to be within $\frac1{1000}$ of the infinite sum is
$$
\sum_{k=1}^{249}(-1)^k\frac1{2k+1}
$$
