Question

when n->infinit:
limit [1/(n^2)+1/(n+1)^2+...+1/(2n)^2]=?
My answer is 0, but the textbook is 1. Am I right?

Answer 1

\[\lim_{n \rightarrow \infty}\frac{1}{n^2} + \frac{1}{n^2+1}+...+\frac{1}{2n^2}\]

Answer 2

No the second item is 1/[(n+1)^2]

Answer 3

Oh Okay

Answer 4

\[\lim_{n \rightarrow \infty}\frac{1}{n^2} + \frac{1}{(n+1)^2}+...+\frac{1}{(2n)^2}\]

Answer 5

It is zero ...right?

Answer 6

Looks zero to me too.

Answer 7

wait satellite is here

Answer 8

not sure it is zero

Answer 9

Thank you for your confirm..

Answer 10

but it is not obvious to me what it is. requires some work yes?

Answer 11

not sure it is zero

Answer 12

I use squeez theorem let Sn=this series then:
1/4n = [1/(2n)^2+1/(2n)^2+...1/(2n)^2]< Sn <[1/n^2+1/n^2+...1/n^2]=1/n

Answer 13

then n-> infinit 0<Sn<0. SO Sn=0 ? But the Squeeze Theorem need 0<=Sn<=0 ...
SO I am not sure

Answer 14

Or just compute the limit for each item, then 0+0+0...=0

Answer 15

yeah looks good to me!

Answer 16

no your first method looks good. you have
\[\frac{1}{n^2}+...+\frac{1}{(2n)^2}<\frac{1}{n^2}+...+\frac{1}{n^2}=\frac{1}{n}\]

Answer 17

be careful adding infinite number of things that tend to zero. not really a method. i like your answer above

Answer 18

you type the equation by code. Is there any help so I can learn it ?

Answer 19

Yes, thank you for your reminding "be careful adding infinite number of things that tend to zero".

Answer 20

Use \ [ without the space to start an equation.
\ ] is used to close an equation. Within the equation \frac{}{} is used to display a fraction. ^{} is used for exponents.