Show that $a_n=\sum\limits_{k={n+1}}^{2n+1}\frac{1}{k}$ converges to $\frac{1}{2}$ if $n \to \infty$

Question

Could you check my calculation please, I got the following result:

$1-\frac{1}{2}\frac{n}{n+1}\lt\sum_{k={n+1}}^{2n+1}\frac{1}{k}\lt1$ where $n=1,2 ....\to\infty$

Answer 1

If you enjoy harmonic numbers $$a_n=\sum\limits_{k={n+1}}^{2n+1}\frac{1}{k}=H_{2 n+1}-H_n$$ Now, using the asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ use it and continue with Taylor expansion for large values of $n$ to get $$a_n=\log (2)+\frac{1}{4 n}-\frac{3}{16 n^2}+O\left(\frac{1}{n^3}\right)$$

Answer 2

It does not converge to $1/2$. By approximating the sum by definite integrals (see here for more details), $$ \log\Bigl(\frac{2(n+1)}{n+1}\Bigr)=\int_{n+1}^{2n+2}x^{-1}dx\le\sum_{k=n+1}^{2n+1}k^{-1}\le\int_n^{2n+1}x^{-1}dx=\log\Bigl(\frac{2n+1}n\Bigr). $$ Hence, $\sum_{k=n+1}^{2n+1}k^{-1}\to\log 2$ as $n\to\infty$.

Answer 3

This is the Riemann sum: $$\lim_\limits{n\to\infty} \sum_{k=1}^{n+1} \frac{1}{n+k}=\lim_\limits{n\to\infty}\frac1n \sum_{k=1}^{n+1}\frac{1}{1+\frac{k}{n}}=\int_0^1 \frac{1}{1+x}dx =\ln (1+x) \bigg{|}_0^1=\ln 2.$$