Question

If n is a positive integer, then lim (n-->infinity) (1/n) [1/(1+(1/n)) + 1/(1+(2/n))+...+1/(1+(n/n)] is?
the above can be expressed as
a.) integral from 0 to 1 of (1/x) dx
b.) integral from 1 to 2 of (1/(x+1))dx
c.) integral from 1 to 2 of (x)dx
d.) integral from 1 to 2 of (2/(x+1))
e.) integral from 1 to 2 of (1/x)

please show me how u got the answer thanks a lot!

Answer 1

Been a while since I did calc. This looks a lot like what they use when they go over the Riemann Sum definition of an integral. Is that where you are at in calc?

Answer 2

might be easier to read in latex

Answer 3

Yah. I was thinking that myself.

Answer 4

1/n represents (b-a)/n which is the interval from a to b

Answer 5

they all have intervals of 1, so that doesnt narrow it down

Answer 6

\[\lim_{n\to inf}~ \sum_{i=1}^{n}\left(\frac{1}{1+\frac in}\right)\frac 1n\]
maybe?

Answer 7

\[Left~side~sums:\lim_{n\to inf}~\sum_{i=0}^{n-1}~f(a+i\frac{b-a}{n})\frac{b-a}{n}\]
\[Right~side~sums:\lim_{n\to inf}~\sum_{i=1}^{n}~f(a+i\frac{b-a}{n})\frac{b-a}{n}\]

Answer 8

i notice some are missing a dx, if that correct, then those can be eliminated for improper format

Answer 9

c doesnt have division in it, so id scratch that

and itd be a toss up between a and b

Answer 10

when n=0, we have 1/1, and not 1/0 soo im leaning towards b

Answer 11

when x = a + i/n

1/(x) = 1/(0+i/n)

there is also a b version

f (b-i(b-a)/n) , for i = 0 to n-1

Answer 12

1/(1+x) = 1/(1+0+1/n) seems to match best tho

Answer 13

lol, im sticking with B :) good luck tho