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Prove that $$\sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k} > \frac{1}{2}.$$

I've tried a lot of things (mainly induction) without much result. The only hint I was given was that if $a \leq c_k \leq b$ for k = 1, 2, ..., n then $na \leq \sum\limits_{k=1}^{n} c_k \leq nb$. I've failed to use that or any other way for this proof. How to prove that?

gt6989b
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DuckHunterZx
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3 Answers3

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You have $$\sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k} \geq \sum_{k=2^n+1}^{2^{n+1}} \frac{1}{2^{n+1}} = \frac{2^{n+1}-2^n}{2^{n+1}} = \frac{1}{2}$$

TheSilverDoe
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  • Could you please explain why the inequality holds and how you got the last part? I am new to this. Thanks. – DuckHunterZx Aug 28 '20 at 13:04
  • I bounded each of the $1/k$ by the smallest of all of them, which is $1/2^{n+1}$ ; then you have to sum $(2^{n+1}-2^n)$ times the constant term $1/2^{n+1}$, which gives you $\frac{2^{n+1}-2^n}{2^{n+1}}$ ; finally, you just have to simplify the fraction. – TheSilverDoe Aug 28 '20 at 13:19
  • Thank you very much, I think I got it. A few more questions before I mark it as correct: Don't we have upper limit - lower limit times the constant term? Wouldn't that be $2^{n+1} - 2^n - 1 $ times 1/k ? Also if i pick the largest 1/k i can find the upper bound similarly? – DuckHunterZx Aug 28 '20 at 13:30
  • For your first question : be careful with the number of terms in your sum : for example, if you sum from $0$ to $10$, you have $11$ terms ! Generally, if you sum from $p$ to $q$, you have $q-p+1$ terms (and not simply $q-p$). For your second question : yes, you may have an upper bound for the sum (but it will depend on $n$). – TheSilverDoe Aug 28 '20 at 13:33
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$$S_n=\sum_{k=2^n+1}^{2^{n+1}} \frac{1}{k} =H_{2^{n+1}}-H_{2^n}$$

Using the asymptotics $$H_p=\gamma +\log \left(p\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ $$S_n=\log (2)-2^{-2 (n+2)} \left(2^{n+2}-1\right)+\cdots$$

So, $S_n$ is an increasing function which for $n=0$ is $$S_0=\log (2)-\frac{3}{16} \approx 0.505647$$ while the exact value is $S_0=\frac 12$

Claude Leibovici
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We have that

$$n=1 \implies \sum\limits_{k=2^n+1}^{2^{n+1}}\frac1k =\frac13+\frac14>2\cdot\frac14=\frac12$$

$$n=2\implies\sum\limits_{k=2^n+1}^{2^{n+1}}\frac1k =\frac15+\frac16+\frac17+\frac18>4\cdot\frac18=\frac12$$

$$\ldots$$

and then by induction

$$\sum\limits_{k=2^n+1}^{2^{n+1}}\frac1k>2^n\cdot\frac1{2^{n+1}}=\frac12$$

This is exactly the way we prove divergence of harmonic series by condensation.

user
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