The limit of truncated sums of harmonic series, lim\lim\limits_{k\to\infty}\sum_{n=k+1}^{2k}{\frac{1}{n}}

What is the sum of the ‘second half’ of the harmonic series?

\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?

More precisely, what is the limit of the above sequence of partial sums?

Answer

Rewriting the sum as

\sum_{n=k+1}^{2k}\frac1n=\sum_{n=k+1}^{2k}\frac1k\cdot\frac1{n/k}

allows us to identify this as a Riemann sum related to the definite integral
\int_1^2\frac1x\,dx=\ln 2.
To see that, divide the interval [1,2] to k equal length subintervals, and evaluate the function f(x)=1/x at the right end of each subinterval. When k\to\infty, the Riemann sums will then tend to the value of this definite integral.

Attribution
Source : Link , Question Author : Daniel Pietrobon , Answer Author :
Jyrki Lahtonen

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