Question

Prove that \(\lim_{N\rightarrow \infty}N\sum_{n=N}^{\infty}\frac{1}{n^2}\) exists.

Answer 1

why prove question draw attention lol

Answer 2

questions*

Answer 3

Because it is interesting question

Answer 4

my first guess, which may be wrong, is to prove that this is cauchy sequence

Answer 5

Since \(N\rightarrow\infty\), N>1. Moreover, \(\frac{1}{n^2}\ge0 \) for all \(n\in N\).
Hence, we have \(0\le\sum_{n=N}^{\infty}\frac{1}{n^2}\le\sum_{n=1}^{\infty}\frac{1}{n^2}\).
Thus, \(0\le N\sum_{n=N}^{\infty}\frac{1}{n^2}\le N\sum_{n=1}^{\infty}\frac{1}{n^2}\).

But then, I am not sure how I can proceed from here, or if I am on the right track...

Answer 6

@geerky42, I agree lol

Answer 7

If it is a Cauchy sequence, then the series converges. But what's next?

Answer 8

i have no idea how to show it is cauchy, that was just my guess as a strategy

Answer 9

Does "it" refer to \(N\sum_{n=N}^{\infty}\frac{1}{n^2}\) or \(\frac{1}{n^2}\) ?

Answer 10

if you are asking me i would say "it" refers to \(N\sum_{n=N}^{\infty}\frac{1}{n^2}\)

Answer 11

you are working too hard

Answer 12

oh damn

Answer 13

try an integral test

Answer 14

don't tell me it is l'hopital or something that simple

Answer 15

We haven't learnt it in our class yet D:

Answer 16

you haven't done the integral test?

Answer 17

Not yet. Only Cauchy, Comparison, Ratio, Root, and Alternating Series Test.

Answer 18

well an integral test is derived from the comparison test

Answer 19

\[\sum_{n=N}^\infty\frac{1}{n^2}\le \int\limits_{N}^{\infty}\frac{1}{x^2}dx =\frac{1}{N}\]

Answer 20

yeah, it is one
damn

Answer 21

\[N\sum_{n=N}^\infty\frac{1}{n^2}\le N\int\limits_{N}^{\infty}\frac{1}{x^2}dx =N\frac{1}{N}=1\]

Answer 22

not real happy though...still missing something

Answer 23

need to show that \(\displaystyle N\sum_{n=N}^\infty\frac{1}{n^2}\) as a sequence in N is monotone

Answer 24

it is monotone. I can show the difference \(a_{N+1}-a_N\) is negative

Answer 25

bounded monotone sequences converge

Answer 26

i should have

\[N\sum_{n=N}^\infty\frac{1}{n^2}\le N\int\limits_{N-1}^{\infty}\frac{1}{x^2}dx =N\frac{1}{N-1}=\frac{1}{1-\frac{1}{N}}\]

Answer 27

Not happy at all...
\[a_{N+1}-a_{N} = (N+1)\sum_{n=N+1}^{\infty}\frac{1}{n^2}-N\sum_{n=N}^{\infty}\frac{1}{n^2} = N(\sum_{n=N+1}^{\infty}\frac{1}{n^2} - \sum_{n=N}^{\infty}\frac{1}{n^2}) + \sum_{n=N+1}^{\infty}\frac{1}{n^2}\]\[=-\frac{1}{N}+\sum_{n=N+1}^{\infty}\frac{1}{n^2}\]
Well...

Answer 28

\[1= N\int\limits_{N}^{\infty}\frac{1}{x^2}dx \]

\[\le N\sum_{n=N}^\infty\frac{1}{n^2}\le N\int\limits_{N-1}^{\infty}\frac{1}{x^2}dx =N\frac{1}{N-1}=\frac{1}{1-\frac{1}{N}}\]

by the squeze theorem

\[1\le \lim_{N\to\infty}N\sum_{n=N}^\infty\frac{1}{n^2}\le\lim_{N\to\infty}\frac{1}{1-\frac{1}{N}}=1\]

the limit is 1

Answer 29

using what I have above you don't need to show the sequence is monotone.

Answer 30

so not only does it converge we know the limit now

Answer 31

We are not expected to use integral test though

Answer 32

I didn't technically. I used the comparison test

Answer 33

the integral test just tells you if a series converse or not. It doesn't give you bounds