Question

Find the sum using the partial sums. as n approaches infinity..
a_n=1/(n+1) - 1/(n+2)

Answer 1

it wants the sum though...however that means it will converge

Answer 2

yes you are correct didn't read question properly

Answer 3

does he want this?:
\[\sum_{n = 0}^{\infty}a_n\]
?
its a telescoping sum.

Answer 4

looks like it
im rusty at these..so i'll let someone else take over

Answer 5

I believe the first one converges to 1 then the second converges to 1/2.

Therefore 1-1/2= 1/2

Answer 6

I used the integral test

Answer 7

If i write out the first n terms i would get:
\[(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{n+1}-\frac{1}{n+2})\]
all the terms in the middle will cancel out, leaving:
\[1-\frac{1}{n+2}\]
So basically you want:
\[\lim_{n\rightarrow \infty}1-\frac{1}{n+2} = 1\]

Answer 8

Wait are you sure the answer is 1 joemath? i dubbed check my work with wolfram and it also said 1/2

Answer 9

very sure.
the integral test doesnt tell you the value of S_n, it just lets you know if it converges or diverges. Since you got an answer using the integral test, that lets you know it converges, but not to what.

Answer 10

it might be that you are starting at n=0 instead of n=1

Answer 11

...or that LOLOLOL

Answer 12

yup thats it dumb cow

Answer 13

i did 1 to infinity

Answer 14

5 medals for you.

Answer 15

i did 0 to infinity.

Answer 16

I'm still unsure which one to use though

Answer 17

was i right in assuming 1 to infinity or is it 0 to infinity

Answer 18

it wasnt specified. so we are both right unless the poster says otherwise.

Answer 19

ahhh ok then

Answer 20

its not like either of us are wrong lol.

Answer 21

ha ha no i didn't no if these question implicitly are 0 to infinity and i just forgot

Answer 22

i really depends on the function of n. if it had been something with 1/n in it, i would have started at 1, since n = 0 isnt defined.

Answer 23

it*

Answer 24

right good point.