For any $n ≥ 5,$ the value of $1+ \frac{1}2 + \frac{1}3+···+\frac{1}{2^n −1}$ lies between

Question

QUESTION: For any $n ≥ 5,$ the value of $$1+ \frac{1}2 + \frac{1}3+···+\frac{1}{2^n −1}$$ lies between

$(A)$ $0$ and $\frac{n}2$

$(B)$ $\frac{n}2$ and $n$

$(C)$ $n$ and $2n$

$(D)$ none of the above.

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MY APPROACH: This is what I did-

We do our calculation considering $n=5$.

Call the given series $S$.

First let us compare the series $S$ with the series $A$ where the denominator of every term of $S$ is changed with the next higher power of $2$ as: $$A=\frac{1}2+\frac{1}2+\frac{1}4+\frac{1}4+\frac{1}8+\frac{1}8+\frac{1}8+\frac{1}8+...+\frac{1}{32}$$ which on grouping gives us-$$\frac{5}2$$ therefore, $A=2.5$

Again, I consider the series $B$ is replaced with a lower power of $2$ which goes like- $$1+\frac{1}2+\frac{1}2+\frac{1}4.4+\frac{1}8.8+... +\frac{1}{16}$$ which gives $B=5$.

Now observe that $A<S<B$. So the value of $S$ must lie in $[2.5,5]$.

So the answer should be $\frac{n}2$ to $n$.

Am I correct? How do I prove it rigorously?

Also, since there is no closed formula for the summation of harmonic progressions, how do I calculate the value of the sum (if it was asked for some large value of $n$)

Can anyone please help me out? Thank you in advance.

Answer 1

Divide the sum into blocks with denominators running from $2^m$ through $2^{m+1}-1$, where $0\le m<n$:

$$\begin{align*} \sum_{k=1}^{2^n-1}\frac1k&=\underbrace{1}+\underbrace{\frac12+\frac13}+\underbrace{\frac14+\frac15+\frac16+\frac17}+\ldots+\underbrace{\frac1{2^{n-1}}+\ldots+\frac1{2^n-1}}\\ &=\sum_{m=0}^{n-1}\sum_{k=2^m}^{2^{m+1}-1}\frac1k\;.\end{align*}$$

The inner summation has $2^m$ terms, so we have

$$\frac12=2^m\cdot\frac1{2^{m+1}}<\sum_{k=2^m}^{2^{m+1}-1}\frac1k\le 2^m\cdot\frac1{2^m}=1\;:$$

each term is larger than $\frac1{2^{m+1}}$, and each term is less than or equal to $\frac1{2^m}$. And the outer summation has $n$ terms, so

$$\frac{n}2<\sum_{k=1}^{2^n-1}\frac1k\le n\;.$$

It is possible to get good approximations to the harmonic numbers quite easily. For instance

$$H_n\sim \ln n+\gamma+\frac1{2n}-\frac1{12n^2}+\frac1{240n^4}\;,$$

where $\gamma\approx0.5772156649$ is the Euler-Mascheroni constant.