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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?

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  • $\begingroup$ If the series fails to satisfy the second condition, you can only conclude that this test does not apply; you can't conclude that the series diverges. $\endgroup$ – user84413 Dec 3 '14 at 15:03
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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.

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  • $\begingroup$ That is also what I thought... But if $u_{k+1} > u_k$ for all $k$ is impossible, then what is the point of the second condition? $\endgroup$ – elDin0 Dec 2 '14 at 15:48
  • $\begingroup$ The second condition says that $u_{k+1}\le u_k$ for all $k$. One can get convergence with weaker conditions, but as the example above shows, if we drop Condition 2 we may have divergence. $\endgroup$ – André Nicolas Dec 2 '14 at 16:06
  • $\begingroup$ If you piecewise define a function, sure. But Im pretty sure conditions 2 and 3 are contradictory for any smooth function that is everywhere real. Maybe Im wrong? $\endgroup$ – CogitoErgoCogitoSum Dec 3 '14 at 1:28
  • $\begingroup$ @CogitoErgoCogitoSum: Using bump functions (technical term!) you can get this kind of behaviour with infinitely differentiable functions. $\endgroup$ – André Nicolas Dec 3 '14 at 2:37
  • $\begingroup$ I see. That is cool. Thanks. Though the wikipedia article on bump functions still shows a piece-wise defined function as an example, which is what I said in my last comment. Can you give an example that isnt piecewise defined? $\endgroup$ – CogitoErgoCogitoSum Dec 3 '14 at 2:55
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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.)

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