Using Riemann integral

Calculus Level 3

$\large \lim_{n\to\infty} \sum_{k=n}^{k=np} \dfrac1k$

If $p$ is a positive integer, compute the limit above.

\ln 2

\ln (p!)

\ln p

1 solution

Aareyan Manzoor
Mar 6, 2016

first, $\lim_{n\to\infty}\left(\left(\sum_{k=1}^n \dfrac{1}{k}\right)-\ln(n)\right)=\gamma$ so, $\lim_{n\to\infty}\left(\sum_{k=n}^{np} \dfrac{1}{k}\right)=\lim_{n\to\infty}\left(\sum_{k=1}^{np} \dfrac{1}{k}-\sum_{k=1}^{n-1} \dfrac{1}{k}\right)\\=\lim_{n\to\infty}\left(\gamma+\ln(np)-(\gamma+\ln(n-1))\right)= \ln(p)$

Using Riemann integral

1 solution

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