Question

evaluate the limit by first recognizing the sum as a rieman sum on the interval 0 to 1.
lim as n approaches infinity, (1/sqrt(n))((1/sqrt(n+2)+(1/sqrt(n+3)...+1/sqrt(n+n)

Answer 1

\[\int_0^1\frac{1}{\sqrt{x+1}}dx\] is my guess

Answer 2

i am assuming you are missing a plus sign at the beginning

Answer 3

no its listed a a reimann sums
and for the other problems we had to find a F(x)

Answer 4

(1/sqrt(n)) is delta x

Answer 5

ok so if
\[\Delta x=\frac{1}{\sqrt{n}}\] then that means
\[f(\frac{1}{\sqrt{n}})=\frac{1}{\sqrt{n+1}}\] and
\[f(\frac{2}{\sqrt{n}})=\frac{1}{\sqrt{n+2}}\] ektc

Answer 6

what the heck is this integral???

Answer 7

or am i completely on the wrong track...

Answer 8

seems like i was off by 1
\[f(\frac{1}{\sqrt{n}})=\frac{1}{\sqrt{n+2}}\] i must be doing something wrong

Answer 9

is this the problem

\[\frac{1}{\sqrt{n}}\left[\sum_{i=2}^{n}\frac{1}{\sqrt{n+i}}\right]\]

Answer 10

i think so yes, and the question is what integral is this

Answer 11

on the interval [0,1] and letting n go to infinity
maybe i am supposed to be thinking
\[\Delta x=\frac{1}{\sqrt{x}}\] and then find x? i am totally at sea here

Answer 12

\[=\frac{\sqrt{n}}{n}\left[\sum_{i=2}^{n}\frac{1}{\sqrt{n+i}}\right]\]

\[=\frac{1}{n}\left[\sum_{i=2}^{n}\frac{\sqrt{n}}{\sqrt{n+i}}\right]\]

\[=\frac{1}{n}\left[\sum_{i=2}^{n}\frac{1}{\frac{\sqrt{n+i}}{\sqrt{n}}}\right]\]

\[=\frac{1}{n}\left[\sum_{i=2}^{n}\frac{1}{\sqrt{1+\frac{i}{n}}}\right]\]

Answer 13

wow so my initial guess was correct?

Answer 14

yes

Answer 15

second luckiest guess here. earlier this summer someone posted a series and said "what is this?" as a challenge and i wrote
\[\frac{e^2}{4}\] as a joke and it was right.

Answer 16

nice

Answer 17

i like to think of it as "intuition" but it really was random