Question

Evaluate::
\[ \huge \lim_{k \rightarrow \infty} \sum_{n = k+1}^{2k} {1 \over n}\]

Answer 1

ln 2

Answer 2

http://www.wolframalpha.com/input/?i=lim+k-%3E+infinity+summation+n%3D1%2Bk+to+2k+of+1%2Fn

no you are not mukushla at least according to wolfram alpha

Answer 3

ln(2) ! yes !

Answer 4

how did you get that just wondering this is too advanced for me probably so I don't know why I bother asking

Answer 5

\[\int\limits_{0}^{1}\frac{1}{x+1}dx\] @mukushla

Answer 6

sorry i was right first time...

Answer 7

log 2

Answer 8

would the answer be a function of \(k\)?

Answer 9

this is pretty unintuitive.

Answer 10

k -> inf

Answer 11

i see the limit now,

Answer 12

still ... this is pretty unintuitive... atleast for me when i saw it for the first time.

Answer 13

can we break the sum into two sums?

Answer 14

\[ \lim_{k \rightarrow \infty} \sum_{n = k+1}^{2k} {1 \over n}= \lim_{k \rightarrow \infty} (\frac{1}{1+k}+\frac{1}{2+k}+...+\frac{1}{2k})\\=\lim_{k \rightarrow \infty} \frac{1}{k}(\frac{1}{1+\frac{1}{k}}+\frac{1}{1+\frac{2}{k}}+...+\frac{1}{1+\frac{k}{k}})=\int\limits_{0}^{1} \frac{1}{1+x} dx=\ln 2 \]

Answer 15

yep ... i believe we can. but this method can be quite difficult ...

Answer 16

\[\sum_{n=k+1}^{2k}=\sum_{n=0}^{2k}-\sum_{n=0}^{k+1}\]

Answer 17

this is related to http://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant

Answer 18

*this method
the other is a lot simpler.

Answer 19

or
http://en.wikipedia.org/wiki/Riemann_integral

Answer 20

Riemann sums is easier also
\[ \sum_{n = k+1}^{2k}{1 \over k{ n \over k}} = \int_1^2 {}1 {\over x } dx\]

Answer 21

*1/x dx

Answer 22

hmm, i dont get it