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How to show that this limit $\lim_{n\rightarrow\infty}\sum_{k=1}^n(\frac{1}{k}-\frac{1}{2^k})$ is divergent?

I applied integral test and found the series is divergent. I wonder if there exist easier solutions.

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Hint: one piece is convergent while the other diverges. –  1015 Jul 3 '13 at 13:24
    
So, Is this enough to say this series is divergent? –  mtm Jul 3 '13 at 13:25
1  
Yes: if $\sum a_n$ diverges and $\sum b_n$ converges, then $\sum a_n-b_n$ diverges, as well as $\sum a_n+b_n$. Because if $\sum c_n$ and $\sum a_n$ both converge, $\sum c_n+a_n$ converges. –  1015 Jul 3 '13 at 13:29
    
@julien, thanks. –  mtm Jul 3 '13 at 13:33

2 Answers 2

up vote 2 down vote accepted

We can estimate. Note that $2^k \ge 2k$, and therefore $$\frac{1}{k}-\frac{1}{2^k} \ge \frac{1}{k}-\frac{1}{2k}=\frac{1}{2k}.$$

It is a familiar fact that $\sum \frac{1}{2k}$ diverges. Thus by Comparison so does our series.

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Each partial sum of your series is the difference between the partial sum of the harmonic series, and the partial sum of the geometric series. The latter are all bounded by 1. Since the harmonic series diverges, your series does also.

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