Question

Find the limit
$\displaystyle \sf \lim_{n \to \infty } \bigg( \frac{1}{n} + \frac{1}{n + 1} + \frac{1}{n + 2} + \cdot \cdot\cdot \cdot\cdot \cdot + \frac{1}{kn} \bigg)$
where k is a positive is integer, hence calculate
${ \displaystyle \frac{1}{1000} + \frac{1}{1001} + \frac{1}{1002} + \cdot \cdot\cdot \cdot\cdot \cdot + \frac{1}{5000} }$
​

Answer 1

Answer:

this is your answer

hshdgamsowbs

Answer 2

Answer:

Step-by-step explanation:

$\displaystyle \sf \lim_{n \to \infty } \bigg( \frac{1}{n} + \frac{1}{n + 1} +\frac{1}{n+2}+\ldots+\frac{1}{kn})\\\\= \displaystyle \sf \lim_{n \to \infty } \bigg( \frac{1}{n+0} + \frac{1}{n + 1} +\frac{1}{n+2}+\ldots+\frac{1}{n + (k-1)n})\\\\= \displaystyle \sf \lim_{n \to \infty } \bigg(\frac{1}{n} + \sum_{r=1}^{(k-1)n} \bigg(\frac{1}{n+r}\bigg)\bigg)\\\\= 0 + \displaystyle \sf \lim_{n \to \infty } \sum_{r=1}^{(k-1)n} \frac{1}{n} \bigg(\frac{n}{n+r}\bigg)~(\because~\frac{1}{\infty}\text{ tends to } 0)$

$=\displaystyle \sf \lim_{n \to \infty } \sum_{r=1}^{(k-1)n} \frac{1}{n} \bigg(\frac{1}{1+\frac{r}{n}}\bigg)\\= \int\limits^{(k-1)}_0 {\frac{1}{1+x}} \, dx~ ~~(~\because~\displaystyle \sf \lim_{n \to \infty } \sum_{r=1}^{pn} \frac{1}{n} f(\frac{r}{n}) = \int\limits^p_0 {f(x)} \, dx ~) \\\\= (\log(1+x))^{k-1} _{0}\\= \log k -\log1\\= \log k$

PART 2:

$\frac{1}{1000}+\frac{1}{1001} +\ldots+\frac{1}{5000}$

Here, n = 1000 and kn = 5000

or, k = 5000/n = 5

So, the required sum is $\log 5$