Alternating Series Test: example which meets all but the decreasing condition?

Question

Leibniz's Alternating Series Test

The series $\sum (-1)^{k-1} u_k$ converges if:

1. $u_k \geq 0$
2. $u_{k+1} \leq u_k$
3. $u_k\rightarrow0$ as $k\rightarrow\infty$

I need to find an alternating series which diverges because it fails to satisfy the second condition (i.e. $u_{k+1} > u_k$). Can anybody help me find such a series?

Answer 1

Hint: Let $u_k=\frac{1}{k}$ when $k$ is odd, and let $u_k=\frac{1}{2^k}$ when $k$ is even.

Show that the series $\sum_1^\infty (-1)^{k+1} u_k$ does not converge.

Remark: Note that in the example above we have $u_{k+1}\gt u_k$ for infinitely many $k$. One cannot do better, since if $u_{k+1}\gt u_k$ for all $k$, then Condition 3 is violated.

Answer 2

You can take the series

$\displaystyle\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\frac{1}{2}(3+(-1)^{k-1})}{\frac{1}{4}(2k+1+(-1)^{k-1})}=\frac{2}{1}-\frac{1}{1}+\frac{2}{2}-\frac{1}{2}+\frac{2}{3}-\frac{1}{3}+\frac{2}{4}-\frac{1}{4}+\cdots$

(Notice that $\dfrac{1}{n}<\dfrac{2}{n+1}$ for $n>1$, and

that the even partial sums of the series are the partial sums of the harmonic series.)