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Anonymous Tue 29 Mar 13:32:58 2022 No.14346475 View Reply Original Report

Quoted By: >>14346481 >>14346483 >>14346491 >>14346499 >>14346595 >>14347186

Can you solve it?

Anonymous

Anonymous Tue 29 Mar 2022 13:34:28 No.14346481 Report

Quoted By:

>>14346475
no

Anonymous

Anonymous Tue 29 Mar 2022 13:35:21 No.14346483 Report

Quoted By:

>>14346475
yes

Anonymous

Anonymous Tue 29 Mar 2022 13:37:10 No.14346491 Report

Quoted By: >>14346498

>>14346475
Seems to be 0 as $\ln (n) \in O(n^2)$ as $n \rightarrow \infty$ so the sum just becomes 0 + 0 + 0 ....

Anonymous

Anonymous Tue 29 Mar 2022 13:37:20 No.14346492 Report

Quoted By: >>14346581 >>14346604 >>14346647 >>14346669 >>14347066

$(n^2 + \ln n) \xrightarrow{n\rightarrow\infty} n^2 \Rightarrow x_n \xrightarrow{n\rightarrow\infty} \frac{1}{n}$

The sum diverges.

Anonymous

Anonymous Tue 29 Mar 2022 13:38:56 No.14346498 Report

Quoted By: >>14346524

>>14346491
Yea except you still have positive terms, just take the first partial sum for example

Anonymous

Anonymous Tue 29 Mar 2022 13:38:57 No.14346499 Report

Quoted By: >>14346520

>>14346475
it's 1
proof: i plugged a big number into my calculator.

Anonymous

Anonymous Tue 29 Mar 2022 13:46:03 No.14346520 Report

Quoted By:

>>14346499
Based retard and calculator-pilled.

Anonymous

Anonymous Tue 29 Mar 2022 13:46:28 No.14346524 Report

Quoted By: >>14346527

>>14346498
All partial sums tend to 0 as n tends to infinite, all partial sums are a rational function where the numertor is O(denominator) so it converges to 0.
Are you sure the n^2 in the denominator is static? Is in:
$x_n = \sum_{i = 1}^{n} \frac{1}{\sqrt{i^2 + ln(n)}}$

Instead of
$x_n = \sum_{i = 1}^{n} \frac{1}{\sqrt{n^2 + ln(n)}}$

Anonymous

Anonymous Tue 29 Mar 2022 13:48:03 No.14346527 Report

Quoted By:

>>14346524
Fuck, meant ln(i) in both cases

Anonymous

Anonymous Tue 29 Mar 2022 14:10:43 No.14346566 Report

Quoted By: >>14346575 >>14346606

$\frac{1}{\sqrt{n^2+\ln k}} = \frac1n + O\left( \frac1{n^3} \ln k \right)$ so 1

Anonymous

Anonymous Tue 29 Mar 2022 14:14:40 No.14346575 Report

Quoted By:

>>14346566
> 1
are you really sure about that anon.

Anonymous

Anonymous Tue 29 Mar 2022 14:19:35 No.14346581 Report

Quoted By:

>>14346492
oof, its okay anon, we all make mistakes

Anonymous

Anonymous Tue 29 Mar 2022 14:26:18 No.14346595 Report

Quoted By: >>14346606

>>14346475
(1/n)*(1/sqrt(1+ln(n)/n^2))*n <= x_n <= (1/n)*(1/sqrt(1+ln(1)/n^2))*n
1 <= lim <= 1
lim = 1

Anonymous

Anonymous Tue 29 Mar 2022 14:34:19 No.14346604 Report

Quoted By: >>14346647

>>14346492
Since when $\lim_{x \rightarrow +\infty} \ln{x} = 0$ ?
Unless I am tripping hard I see no reason for it to happen...

Anonymous

Anonymous Tue 29 Mar 2022 14:37:20 No.14346606 Report

Quoted By: >>14347065

>>14346566
a bit more reliably:
terms are between $\frac1{\sqrt{n^2+\ln n}}$ and $\frac1n$ so sum is between $\frac{n}{\sqrt{n^2+\ln n}}$ and 1 and $\frac{n}{\sqrt{n^2+\ln n}} = \frac1{\sqrt{1 + \frac1{n^2} \ln n}} \to 1$
I see >>14346595 posted basically the same thing already

Anonymous

Anonymous Tue 29 Mar 2022 15:00:44 No.14346647 Report

Quoted By: >>14346665 >>14347517

>>14346604
It's not. But in comparison to x^2 it's a negligible term. The above post
>>14346492
Is arguing that when n is large, n^2 + ln(n) scales as n^(2). Which is of course true. If you're unconvinced, shift your sum from 1 to infty to the sum from Graham's number to infinity. See what happens

Anonymous

Anonymous Tue 29 Mar 2022 15:17:51 No.14346665 Report

Quoted By: >>14346669

>>14346647
The sum diverges because it tends asymptotically to $\frac{1}{n}$ whose sum is infinite.

Anonymous

Anonymous Tue 29 Mar 2022 15:21:57 No.14346669 Report

Quoted By:

>>14346665
Yes, which is what this post argued
>>14346492

Anonymous

Anonymous Tue 29 Mar 2022 16:54:17 No.14346850 Report

Quoted By:

It converges to 1.

Anonymous

Anonymous Tue 29 Mar 2022 18:29:01 No.14347065 Report

Quoted By:

>>14346606
this
/thread

Anonymous

Anonymous Tue 29 Mar 2022 18:30:03 No.14347066 Report

Quoted By:

>>14346492
based retard

Anonymous

Anonymous Tue 29 Mar 2022 19:23:13 No.14347182 Report

Quoted By: >>14347517

$x_n = \sum_{k=1}^n \frac{1}{\sqrt{n^2 + \ln(k)}} < \sum_{k=1}^n \frac{1}{\sqrt{n^2}} = 1<br />$

$x_n = \sum_{k=1}^n \frac{1}{\sqrt{n^2 + \ln(k)}} > \sum_{k=1}^n \frac{1}{\sqrt{n^2 + n}} = \frac{1}{\sqrt{1 + n^{-1}}} \xrightarrow[n \to \infty]{} 1<br />$

Therefore by the squeeze theorem
$x_n \xrightarrow[n \to \infty]{} 1$

Anonymous

Anonymous Tue 29 Mar 2022 19:25:20 No.14347186 Report

Quoted By:

>>14346475
diverges, each term is greater than 1/[sqrt(2) n]

Anonymous

Anonymous Tue 29 Mar 2022 22:06:25 No.14347517 Report

Quoted By:

>>14346647
Yes, and lnx increases at a rate of 1/x, so as n gets bigger and bigger lnx will increase by a lesser and lesser value.
>>14347182
This one.

Anonymous

Anonymous Wed 30 Mar 2022 06:33:30 No.14348454 Report

Quoted By:

$<br />\sqrt{n^2 + \ln n} < \sqrt{n^2 + 3n^3} = 2n \Rightarrow \frac{1}{\sqrt{n^2 + \ln n}} > \frac{1}{2n}<br />$
So, the sum diverges by the comparison test.

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