What do you get when you sum over the smaller half of the harmonic series?

Question

More specifically, how would you evaluate the below formula? $$\lim_{n\to\infty}\sum_{k=n/2}^{n}\frac{1}{k}$$ I know that the harmonic series starting at any point diverges, but when we limit it in this way, does the series diverge or converge?

If it diverges:

• How might you determine that?

• Is there some $d$ that we can replace with $2$ to make the sequence converge?

If it converges:

• What does it converge to, and how might you determine that?

• The sequence must converge for any $d>2$. Is there a formula for the series generalized for any $d$?

Answer 1

For simplicity let's assume that $n$ is even (if it is odd your sequence has to be modified using integer part of $\frac{n}{2}$). Let's compute $$ \int\limits_{\frac{n}{2}}^{n} \frac{1}{ x} dx = \ln n-\ln(\frac{n}{2}) = \ln 2 $$ Now we have estimations $$ \int\limits_{\frac{n}{2} - 1}^{n} \frac{1}{ x} dx \ge \sum_{k=n/2}^{n}\frac{1}{k} \ge \int\limits_{\frac{n}{2}}^{n + 1} \frac{1}{ x} dx $$ and both estimates obviously tend to $\ln 2$ as $n \rightarrow \infty$.

With arbitrary $d$ replacing $2$ we have $\ln d$ as limit.

Answer 2

Let $H_n = \sum_{k=1}^n 1/k$ denote the $n$-th Harmonic number. It is well-known that $(H_n - \log n)_n$ converges to some constant $\gamma$, called the Euler-Mascheroni constant.

You ask for $(H_n - H_{n/2})_n$. We write it as:

$$(H_n - \log n) - (H_{n/2} - \log (n/2)) + (\log (n) - \log (n/2))$$

The two former terms tend to $\gamma$ thus it remains $\log (n) - \log (n/2)$, which is just $-\log (1/2) = \log 2$.

This works for any $d$.

Answer 3

I thought it might be instructive to present a way forward that does not rely on Riemann sums or properties of Harmonic numbers.

Rather, we rely on only elementary analysis and the Taylor series for $\displaystyle \log(1+x)$. To that end, we now proceed.

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The limit $\lim_{n\to\infty}\sum_{k=n/2}^n \frac1k$ converges. To see this we write

$$\begin{align} \sum_{k=1}^{2m} \frac1k&=\sum_{k=1}^m \frac1{2k}+\sum_{k=1}^m\frac1{2k-1}\\\\\ &=\sum_{k=1}^{m} \frac1{k}+\sum_{k=1}^m\left(\frac1{2k-1}-\frac1{2k}\right)\\\\ \sum_{k=m+1}^{2m}\frac1k&=\sum_{k=1}^m\left(\frac1{2k-1}-\frac1{2k}\right)\\\\ &=\sum_{k=1}^{2m}\frac{(-1)^{k+1}}{k}\tag1 \end{align}$$

Letting $m=n/2$ in $(1)$ yields

$$\lim_{n\to\infty}\left(\sum_{k=n/2}^n\frac1k-\frac1{n/2}\right)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}\tag2$$

The right-hand side of $(2)$ is $\log(2)$ as recognized from the Taylor Series of $\log(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k$ evaluated at $x=1$. And the right-hand side reduces to the limit of interest. Hence, we see that

$$\lim_{n\to\infty}\sum_{k=n/2}^n\frac1k=\log(2)$$

And we are done!

Answer 4

Concerning the limit itself, you already received the good answers.

What I found interesting is the problem of the partial sums for large values of $n$. $$S_{n,d}=\sum_{k=\frac n d}^{n}\frac{1}{k}=H_n-H_{\frac{n}{d}-1}$$ Using, for large $p$, the expansion $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ apply it twice an continue with Taylor series to get $$S_{n,d}=\log (d)+\frac{d+1}{2 n}+\frac{d^2-1}{12 n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the respective roles of $n$ and $d$.