/sci/ - Science & Math » Thread #11619297

38KiB, 680x1005, 7b3.jpg

View Same Google iqdb SauceNAO

Solve this problem i made up

Anonymous Thu 30 Apr 06:39:46 2020 No.11619297 View Reply Original Report

Quoted By: >>11619309 >>11619325 >>11619332 >>11619425 >>11619520

difficulty: fukken easy lmfao

prove that there is no sequence $\{b_n\}$ , $b_n \rightarrow 0$ with $1/n < b_n - b_{n-1}$ .

iq threshold for solve: 60

Anonymous

Anonymous Thu 30 Apr 2020 06:45:10 No.11619307 Report

Quoted By:

this belongs in sqt

Anonymous

Anonymous Thu 30 Apr 2020 06:45:18 No.11619309 Report

Quoted By: >>11619315

>>11619297
Consider the sequence lim x-> infinity ((-1)^x)/x

1/2 - (-1) = 3/2 > 1/2

Anonymous

Anonymous Thu 30 Apr 2020 06:48:04 No.11619315 Report

Quoted By: >>11619321

>>11619309
$\forall n$ low iq nig

Anonymous

Anonymous Thu 30 Apr 2020 06:54:53 No.11619321 Report

Quoted By: >>11619323

>>11619315
Well why didn't you say so in the OP? :^)

Anonymous

Anonymous Thu 30 Apr 2020 06:56:23 No.11619323 Report

Quoted By: >>11619341

>>11619321
ngl p. low iq move from me

Anonymous

Anonymous Thu 30 Apr 2020 06:56:47 No.11619325 Report

Quoted By:

>>11619297
>iq threshold for solve: 60
poor 57 IQ OP can't do his 60 IQ homework

Anonymous

Anonymous Thu 30 Apr 2020 06:58:14 No.11619332 Report

Quoted By: >>11619500

>>11619297
Thinly Veiled Homework Thread

Anonymous

Anonymous Thu 30 Apr 2020 06:59:51 No.11619341 Report

Quoted By:

>>11619323
0/10 not enough rigor

Anonymous

Anonymous Thu 30 Apr 2020 07:56:09 No.11619425 Report

Quoted By:

>>11619297
show that the sequence must be increasing so it must approach 0 from the left and you get contradiction

Anonymous

Anonymous Thu 30 Apr 2020 08:28:34 No.11619500 Report

Quoted By:

>>11619332
literally its not a homework thread. i literally made this up.
src: me lol

Anonymous

Anonymous Thu 30 Apr 2020 08:36:40 No.11619520 Report

Quoted By:

>>11619297
K since u guyz r lo iq fags here's the fukken solution u dumb nigs:

Suppose on the cont. that condition in OP is true for som $\{b_n\}$ . Consider the series $\sum_{n=1}^{\infty}(b_n - b_{n-1})$ . The n-th partial sum is $b_n - b_0$ . Since $b_n /rightarrow 0$ the series converges to $-b_0$ .

Lol but this is impossible since by comparison the series $\sum_{n=0}^{\infty}\frac{1}{n}$ would have to converge. every fag on earth knos the harmonic series diverges so ther u go no such sequence exists.

I thought it was neat lol.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches