1. tricky prob;em of summation

X=1/1001 + 1/1002 + ........ + 1/3001 then..

a) X<1
b) X>3/2
c) 1<X<3/2
d) None of the above

2. Originally Posted by biplab
X=1/1001 + 1/1002 + ........ + 1/3001 then..

a) X<1
b) X>3/2
c) 1<X<3/2
d) None of the above
c)

3. Thanks..but can anybody please hep me in detail solution...

4. Originally Posted by biplab
X=1/1001 + 1/1002 + ........ + 1/3001 then..

a) X<1
b) X>3/2
c) 1<X<3/2
d) None of the above
$1+\frac{1}{2}+...+\frac{1}{n} \approx \ln n$. I can give you an error estimate but you probably are not interested in that. Thus, $1+\frac{1}{2}+...+\frac{1}{3001} \approx \ln 3001$. But $1+\frac{1}{2}+...+\frac{1}{1000}\approx \ln 1000$. This means, $\frac{1}{1001}+...+\frac{1}{3001}\approx \ln 3001 - \ln 1001 = \ln (3001/1001) = 1.09...$.

5. Hello,

Originally Posted by ThePerfectHacker
$1+\frac{1}{2}+...+\frac{1}{n} \approx \ln n$
I'm only curious : how do you get this approximation ? ?

6. Originally Posted by Moo
I'm only curious : how do you get this approximation
Since you ask I would have to give you the error term as well.

Consider the function $f(x) = \frac{1}{x}$ on the interval $[1,n]$ where $n\geq 2$ is an integer.
Then, approximating by Riemann sums,
$\sum_{k=1}^{n-1} \frac{1}{k} \geq \int_1^n \frac{dx}{x} \geq \sum_{k=2}^n\frac{1}{k}$
Thus, $H_n - \frac{1}{n} \geq \log n \geq H_n - 1 \implies |H_n - \log n| \leq \frac{1}{n}$.

This is Mine 94th Post!!!

7. Oh I see, I was looking towards power series and Taylor series ^^'

Thanks