/sci/ - Science & Math

[Anonymous No.4680207]

heres what im trying to do

prove that <span class="math"> \sum_{n=0}^{\infty} \frac{n^{k}}{10^{n}} [/spoiler] converges for any real k

i wanted to do this after proving it converges for k=0 k=1 and k=2, to 1/9, 10/81 and 110/729 respectively

[Anonymous No.4680219]

Pinching theorem is your friend

[Anonymous No.4680295]

>>4680207
ratio test and if it converges to a value below 1 then the series converges
(n+1)^k/(10^(n+1))/(n^k)/(10^n) which equals (n+1)^k/(10(n^k)) which can be written as (1+(1/n))^k/10 now let go to infinity and you get 1^k/10 will always be les than 1 for any k so the series must converge

[Anonymous No.4680298]

>>4680295
>>4680295
I meant let "now let n* go to infinty"

inb4 ungrateful OP

[Anonymous No.4680432]

what's the picture from? is there a larger version?

[Anonymous No.4680447]

>>4680295
so

"if <span class="math"> \lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| = L < 1, then \sum_{n=0}^{\infty} a_n converges [/spoiler] "

<span class="math"> a_n = \frac{n^{k}}{10^{n}}
a_{n+1} = \frac{(n+1)^{k}}{10^{n+1}} = \frac{(n+1)^{k}}{10*10^{n}}

\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \rightarrow \infty} |\frac{\frac{(n+1)^{k}}{10^{n+1}}}{\frac{n^{k}}{10^{n}}}| = \lim_{n \rightarrow \infty} |\frac{(n+1)^{k}}{10n^{k}}| = \lim_{n \rightarrow \infty} |\frac{1}{10}| = \frac{1}{10} = L < 1

so \sum_{n=0}^{\infty} \frac{n^{k}}{10^{n}} converges [/spoiler]

[Anonymous No.4680458]

>>4680432

[Anonymous No.4680467]

>>4680458
thank you

[Anonymous No.4680470]

>>4680295
so

"if <span class="math"> \lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| = L < 1[/spoiler] , then <span class="math">\sum_{n=0}^{\infty} a_n[/spoiler] converges "

<span class="math"> a_n = \frac{n^{k}}{10^{n}} [/spoiler]

<span class="math">a_{n+1} = \frac{(n+1)^{k}}{10^{n+1}} = \frac{(n+1)^{k}}{10*10^{n}}[/spoiler]

<span class="math">\lim_{n \rightarrow \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \rightarrow \infty} |\frac{\frac{(n+1)^{k}}{10^{n+1}}}{\frac{n^{k}}{10^{n}}}| = \lim_{n \rightarrow \infty} |\frac{(n+1)^{k}}{10n^{k}}| = \lim_{n \rightarrow \infty} |\frac{1}{10}| = \frac{1}{10} = L < 1[/spoiler]

so <span class="math">\sum_{n=0}^{\infty} \frac{n^{k}}{10^{n}}[/spoiler] converges

[Anonymous No.4680503]

yeah, that was part one, now i want to find a formula that given k returns the value of that sum

[Anonymous No.4680523]

>>4680503

[Anonymous No.4680559]

>>4680458
awesome game, btw
>>4680503
Derp, according to W|A<div class="math">\sum_0^{infty}\frac{n^x}{10^n}=\Phi\right ( \frac{1}{10},-x,0\left ) </div>where <div class="math">\Phi(z,s,a)=\frac{1}{\Gamma(s)}\int_0^\infty
\frac{t^{s-1}e^{-at}}{1-ze^{-t}}\,dt</div>

[Anonymous No.4680564]

>>4680559
derp<div class="math">\sum_0^{infty}\frac{n^x}{10^n}=\Phi (\frac{1}{10},-x,0)</div>

[Anonymous No.4680770]

>>4680559
cool

[Anonymous No.4680838]

im finding a polynomial <span class="math">p_k(n) = a_kn^{k} + a_{k-1}n^{k-1} + ... + a_2n^{2} + a_1n + a_0 [/spoiler]
such that <span class="math">a_k=\frac{10}{9}[/spoiler], <span class="math">a_{k-1}=\frac{10k}{81}[/spoiler] and the others i haven't figured out yet

it'll make so that <span class="math"> \sum_{n=0}^{\infty} \frac{n^{k}}{10^{n}} = p_k(1)/10 [/spoiler]

[Anonymous No.4680872]

>>4680838
<span class="math"> a_{k-2} = \frac{110(k^{2}-k)}{9^{3}.2} [/spoiler]

[Anonymous No.4680880]

>>4680458
This shit is small, yo
Take that

[Anonymous No.4680894]

You can also think in Vinogradov notation, OP.

<span class="math">n^{k} << 2^n[/spoiler]
as n goes through the positive integers with k held constant

So for a given k, after some finite number of terms, every term of that series is less than (2/10)^n ...

And then you just use the convergence of the geometric series
<span class="math">\sum_{0}^{\infty} (2/10)^n[/spoiler]

[Anonymous No.4680921]

>>4680872
got it

<span class="math"> a_{k-r} = \frac{10k!}{9(k-r)!}.\sum{x=1}^{n}\frac{1}{9^{x}(n-x+1)!(x-1)!} [/spoiler]

[Anonymous No.4680929]

>>4680872
got it

<span class="math"> a_{k-r} = \frac{10k!}{9(k-r)!}.\sum_{x=1}^{n}\frac{1}{9^{x}(n-x+1)!(x-1)!} [/spoiler]

[Anonymous No.4680979]

okay, so

<span class="math"> p_k(n) = \sum_{y=0}^{k} a_yn^{y} = \sum_{r=0}^{k} a_{k-r}n^{k-r} = \sum_{r=0}^k ( n^{k-r} \frac{10k!}{9(k-r)!} \sum_{x=1}^{n}\frac{1}{9^{x}(n-x+1)!(x-1)!} ) [/spoiler]

[Anonymous No.4681004]

<span class="math"> p_k(1) = \sum_{r=0}^k ( 1^{k-r} \frac{10k!}{9(k-r)!} \sum_{x=1}^{1}\frac{1}{9^{x}(1-x+1)!(x-1)!} ) [/spoiler]

= <span class="math"> \sum_{r=0}^{k} ( \frac{10k!}{9(k-r)!} \frac{1}{9} ) = \frac{10}{81} \sum_{r=0}^{k} \frac{k!}{(k-r)!} [/spoiler]

[Anonymous No.4681011]

= <span class="math"> 10k!/81 * sum_{r=0}^k 1/r! [/spoiler]

[Anonymous No.4681022]

= <span class="math"> \frac{10k!}{81} \sum_{r=0}^{k} \frac{1}{r!} [/spoiler]

[Anonymous No.4681041]

lol, my wrong is i used n for a new thing, when it already was something, then i thought they were the same

i switched r for n

[Anonymous No.4681052]

>>4680929
>>4680929

<span class="math"> a_{k-r} = \frac{10k!}{9(k-r)!}.\sum_{x=1}^{r}\frac{1}{9^{x}(r-x+1)!(x-1)!} [/spoiler]

<span class="math"> p_k(n) = \sum_{y=0}^{k} a_yn^{y} = \sum_{r=0}^{k} a_{k-r}n^{k-r} = \sum_{r=0}^k ( n^{k-r} \frac{10k!}{9(k-r)!} \sum_{x=1}^{r}\frac{1}{9^{x}(r-x+1)!(x-1)!} ) [/spoiler]

<span class="math"> p_k(1) = \sum_{r=0}^k ( 1^{k-r} \frac{10k!}{9(k-r)!} \sum_{x=1}^{r}\frac{1}{9^{x}(r-x+1)!(x-1)!} ) = \sum_{r=0}^k ( \frac{10k!}{9(k-r)!} \sum_{x=1}^{r}\frac{1}{9^{x}(r-x+1)!(x-1)!} ) [/spoiler]

[Anonymous No.4681438]

proof that lim m->oo pk(m+1)/10^(m+1) = 0

remembering i've defined 10pk(n) - pk(n+1) = 10n^k
and pk(n) as a polynomial of order k
pk(m+1) = 10m^k-10pk(m) = 10(m^k - akm^k - ak-1m^k-1 -...) but this is not gonna help

given k finite, we have pk(m+1)/10^(m+1) as infinite/infinite, which we can derive over and under because l'hopital, giving a polynomial of order k-1 and log10.10^(m+1)
after deriving pk(m+1) k times we get a constant and after deriving 10^(m+1) k times we get (log10)^k.10^(m+1) which is a constant times infinite, so the limit of const/infinite = 0

[Anonymous No.4681444]

oops, i meant

pk(m+1) = 10pk(m) -10m^k

[Anonymous No.4681526]

>>4680564
but it uses the factorial of a negative number

[Anonymous No.4681550]

lol, my formula for a_{k-r} is wrong

[Anonymous No.4681637]

the best way to define <span class="math"> a_{k-r} [/math is as a recursive sequence

starting with <span class="math"> a_k = 10/9 [/spoiler]

<span class="math"> a_{k-r} = \frac{1}{9} (a_k{k \choose r} + a_{k-1}{{k-1} \choose {r-1}} + ... + a_{k-(r-1)}{{k-(r-1)} \choose {r-(r-1)}}) = \frac{1}{9} \sum_{j=0}^{r-1} a_{k-j}{{k-j} \choose {n-j}} [/spoiler][/spoiler]

[Anonymous No.4681639]

the best way to define <span class="math"> a_{k-r} [/spoiler] is as a recursive sequence

starting with <span class="math"> a_k = 10/9 [/spoiler]

<span class="math"> a_{k-r} = \frac{1}{9} (a_k{k \choose r} + a_{k-1}{{k-1} \choose {r-1}} + ... + a_{k-(r-1)}{{k-(r-1)} \choose {r-(r-1)}}) = \frac{1}{9} \sum_{j=0}^{r-1} a_{k-j}{{k-j} \choose {n-j}} [/spoiler]

[Anonymous No.4681671]

these are wrong, guys

[Anonymous No.4681675]

>>4681052
>>4681022
>>4681004
>>4680979
>>4680929
these are wrong, guys

[Anonymous No.4681962]

>>4681639
with this, i want to prove that <span class="math"> \sum_{n=0}^{infty} \frac{n^k}{10^n} = \frac{p_k(1)}{10} = a_0 [/spoiler]

<span class="math">a_{k-r} = \frac{1}{9} \sum_{x=0}^{r-1} a_{k-x}{{k-x} \choose {r-x}} [/spoiler]

<span class="math">a_{k-r} = a_0[/spoiler]
<span class="math">r=k[/spoiler]

<span class="math">a_0 = a_{k-k} = \frac{1}{9} \sum_{x=0}^{r-1} a_{k-x}{{k-x} \choose {k-x}} = \frac{1}{9} \sum_{x=0}^{r-1} a_{k-x} = \frac{1}{9} (a_k+a_{k-1}+...+a_1}[/spoiler]

<span class="math">p_k(1) = a_k+a_{k-1}+...+a_1+a_0 = (a_k+a_{k-1}+...+a_1) + \frac{1}{9} (a_k+a_{k-1}+...+a_1) = \frac{10}{9} (a_k+a_{k-1}+...+a_1)[/spoiler]

<span class="math">\frac{p_k(1)}{10} = \frac{\frac{10}{9} (a_k+a_{k-1}+...+a_1)}{10} = \frac{1}{9} (a_k+a_{k-1}+...+a_1) = a_0[/spoiler]

[Anonymous No.4681974]

>>4681639
with this, i want to prove that <span class="math"> \sum_{n=0}^{infty} \frac{n^k}{10^n} = \frac{p_k(1)}{10} = a_0 [/spoiler]

<span class="math">a_{k-r} = \frac{1}{9} \sum_{x=0}^{r-1} a_{k-x}{{k-x} \choose {r-x}} [/spoiler]

<span class="math">a_{k-r} = a_0[/spoiler]
<span class="math">r=k[/spoiler]

<span class="math">a_0 = a_{k-k} = \frac{1}{9} \sum_{x=0}^{r-1} a_{k-x}{{k-x} \choose {k-x}} = \frac{1}{9} \sum_{x=0}^{r-1} a_{k-x} = \frac{1}{9} (a_k+a_{k-1}+...+a_1)[/spoiler]

<span class="math">p_k(1) = a_k+a_{k-1}+...+a_1+a_0 = (a_k+a_{k-1}+...+a_1) + \frac{1}{9} (a_k+a_{k-1}+...+a_1) = \frac{10}{9} (a_k+a_{k-1}+...+a_1)[/spoiler]

<span class="math">\frac{p_k(1)}{10} = \frac{\frac{10}{9} (a_k+a_{k-1}+...+a_1)}{10} = \frac{1}{9} (a_k+a_{k-1}+...+a_1) = a_0[/spoiler]

[Anonymous No.4682088]

so far i've:
given two different definitions of <span class="math">p_k(n)[/spoiler] and not shown them to be equal

definition 1: <span class="math">p_k(n)=a_{k}n^k+a_{k-1}n^{k-1} + ... + a{1]n + a_0[/spoiler] such that <span class="math">f(n) = p_k(n)/10^n[/spoiler] and <span class="math">f(n) - f(n+1) = n^k/10^n[/spoiler]

definition 2: <span class="math">p_k(n) = \sum_{r=0}^k ( n^{k-r} \frac{1}{9} \sum_{x=0}^{r-1} a_{k-x}{{k-x} \choose {r-x}}[/spoiler]

for definition 1, i've yet to show that \sum_{n=0}^{\infty} n^k/10^n = p_k(1)/10
but i've show that, if it is true, and if definition 1 equals definition 2, then \sum_{n=0}^{\infty} n^k/10^n also equals a_0

after doing these, ill have to find a better formula for a_0 and then the work is done

[Anonymous No.4682097]

so far i've:
given two different definitions of <span class="math">p_k(n)[/spoiler] and not shown them to be equal

definition 1: <span class="math">p_k(n)=a_{k}n^k+a_{k-1}n^{k-1} + ... + a{1}n + a_0[/spoiler] such that <span class="math">f(n) = p_k(n)/10^n[/spoiler] and <span class="math">f(n) - f(n+1) = n^k/10^n[/spoiler]

definition 2: <span class="math">p_k(n) = \frac{1}{9} \sum_{r=0}^k ( n^{k-r} \sum_{x=0}^{r-1} a_{k-x}{{k-x} \choose {r-x}})[/spoiler] and <span class="math">a_k=10/9[/spoiler]

for definition 1, i've yet to show that \sum_{n=0}^{\infty} n^k/10^n = p_k(1)/10
but i've show that, if it is true, and if definition 1 equals definition 2, then \sum_{n=0}^{\infty} n^k/10^n also equals a_0

after doing these, ill have to find a better formula for a_0 and then the work is done

[Anonymous No.4682100]

so far i've:
given two different definitions of <span class="math">p_k(n)[/spoiler] and not shown them to be equal

definition 1: <span class="math">p_k(n)=a_{k}n^k+a_{k-1}n^{k-1} + ... + a{1}n + a_0[/spoiler] such that <span class="math">f(n) = p_k(n)/10^n[/spoiler] and <span class="math">f(n) - f(n+1) = n^k/10^n[/spoiler]

definition 2: <span class="math">p_k(n) = \frac{1}{9} \sum_{r=0}^k ( n^{k-r} \sum_{x=0}^{r-1} a_{k-x}{{k-x} \choose {r-x}})[/spoiler] and <span class="math">a_k=10/9[/spoiler]

for definition 1, i've yet to show that <span class="math">\sum_{n=0}^{\infty} n^k/10^n = p_k(1)/10[/spoiler]
but i've show that, if it is true, and if definition 1 equals definition 2, then <span class="math">\sum_{n=0}^{\infty} n^k/10^n[/spoiler] also equals a_0

after doing these, ill have to find a better formula for a_0 and then the work is done

[Anonymous No.4682103]

am i aspie as fuck?

[Anonymous No.4682240]

There's a trick for computing the sum based on differentiating under the summation:
<div class="math">
\frac{d}{dr} \sum_{n=1}^\infty n^k r^n
= \sum_{n=1}^\infty n^k \frac{d}{dr} r^n
= \sum_{n=1}^\infty n^k \cdot n r^{n-1}
= \frac{1}{r} \sum_{n=1}^\infty n^{k+1} r^n
</div><div class="math"> \sum_{n=1}^\infty n^{k+1} r^n = r \frac{d}{dr} \sum_{n=1}^\infty n^k r^n </div>(I'm starting from n=1, but it only matters for the first term.)

We start with
<div class="math"> \sum_{n=1}^\infty n^0 r^n = \frac{r}{1 - r} = \frac{1}{1/r - 1} </div>and we expect others to gain additional powers of (1/r-1) in the denominator.

So write it like
<div class="math"> \sum_{n=1}^\infty n^k r^n = \frac{p_k(1/r)}{(1/r-1)^{k+1}} </div>where <span class="math">p_k[/spoiler] is a polynomial.

Then:
<div class="math"> \sum_{n=1}^\infty n^{k+1} r^n = \frac{(k+1) (1/r) p_k(1/r)}{(1/r-1)^{k+2}} - \frac{(1/r) p_k'(1/r)}{(1/r-1)^{k+1}} </div><div class="math"> p_{k+1}(x) = (k+1) x p_k(x) + (1 - x) x p_k'(x) </div>

[Anonymous No.4682322]

the way i'd show it is that for any k there exists a number N such that for any n>N:
n^k < 10 ^ n
(basic proof)
then the sum on n^k / 10^n is bounded by the sum on 1/ 10^ (n-1) which converges (each sum taken from N on, but adding finitely many elements doesnt change convergence) bringing us to QED

[Anonymous No.4682882]

just found something

http://oeis.org/search?q=1%2C10%2C110%2C1410&sort=&language=english&go=Search

sum of n^k/10^n would be b(k+1)/3^2(k+1)