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I am having hard time trying to show that the following series is divergent. Can someone help me please?

$$\sum \frac{(-1)^n}{\log n} b_n $$

where $b_n=\frac{1}{\log n} $ if n is even and $b_n=\frac{1}{2^n}$ if n is odd.

This would be a great help!

liamdalton
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user67035
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    The subseries of odd terms converges by comparison with the geometric series $(1/2)\sum 1/2^{2n}$. So your series has the same nature as the subseries of even terms. Now this one is $\sum 1/(\log (2n))^2$. This diverges by comparison with the harmonic series, for instance. – Julien Mar 16 '13 at 21:29

2 Answers2

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The Idea: The terms with $n$ odd are negative. But they don't help much, since there is an absolute bound $b$ such that the absolute value of the sum of any finite number of them is $\lt b$. This is because of the rapid decay of $\frac{1}{2^n}$.

But $\sum_1^m \frac{1}{\log^2(2k)}$ grows without bound, so can be made arbitrarily large, say larger than $C$ for any given $C$. For it is not hard to show by comparison that the series $\sum \frac{1}{\log^2(2k)}$ diverges.

Thus the partial sum of our series can be made larger than $C-b$ for any $C$.

André Nicolas
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Your series is: $$ \frac{1}{\log{2}}\frac{1}{\log{2}} - \frac{1}{\log{3}}\frac{1}{2^3} + \frac{1}{\log{4}}\frac{1}{\log{4}}-\frac{1}{\log{5}}\frac{1}{2^5}+-\ldots $$ Which is: $$ \sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2 - \sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}} $$ $$ \sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}} < \sum_{2}^{\infty} \frac{1}{2^{2n+1}} = \frac{1}{24} $$

So $$ \sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2 - \sum_{2}^{\infty} \frac{1}{\log{(2n+1)}2^{2n+1}} > \left(\sum_{2}^{\infty}\left(\frac{1}{\log{2n}}\right)^2\right) - \frac{1}{24} =\\ \infty - \frac{1}{24} = \infty $$

Rustyn
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