SOLUTION TO PROBLEM 7

Let $S_N=\sum_{n=0}^{N}a_n$ for $N=0, 1, \ldots$. Then

$$\begin{eqnarray*}S_N&=& \sum_{n=0}^N\left (\frac{1}{4n+1}+\frac{1}{4n+3}-\frac{1... ...(-1)^{m-1}}{m}+\frac{1}{2}\sum_{m=1}^{2N+2}\frac{(-1)^{m-1}}{m}. \end{eqnarray*}$$

Letting $N\rightarrow 0$ we have

$$\begin{eqnarray*}\lim_{N\rightarrow 0}S_N&=& \sum_{m=1}^{\infty}\frac{(-1)^{m-1... ...1}}{m}\\ &=&\frac{3}{2}\sum_{m=1}^{\infty}\frac{(-1)^{m-1}}{m}. \end{eqnarray*}$$

But the alternating harmonic series converges to $\ln 2$. So

$$\sum_{n=0}^{\infty}a_n=\frac{3}{2}\ln 2.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Oct 12 09:35:18 CDT 2001