Summation

[man111]

If [tex]\displaystyle I_{r} = \int^{e}_{1}\frac{1}{(1+\ln x)^n}dx.[/tex] Then value of [tex]\displaystyle \sum^{\infty}_{r=0}(1-rI_{r})[/tex] is

[man111]

If [tex]\displaystyle I_{r} = \int^{e}_{1}\frac{1}{(1+\ln x)^r}dx.[/tex] Then value of [tex]\displaystyle \sum^{\infty}_{r=0}(1-rI_{r})[/tex] is

[Davids]

First we're going to solve for $I_r$:
$I_r = \int\limits_{1}^{e}(1 + lnx)^{-r}dx$
We substitute $u = 1 + lnx$
$\Rightarrow x = e^{u - 1}$
$\Rightarrow du = \frac{1}{x}dx \Rightarrow dx = e^{u - 1}du$

We shall also not forget to switch boundaries:
$\Rightarrow I_r = \int\limits_1^2u^{-r}.e^{u - 1}du$

Now we employ the DI method for integration by parts:
$$
\begin{array}{ c| c | c }
&D&I \\
\hline
+&u^{-r}&e^{u - 1} \\
\hline
-&-r.u^{-r-1}&e^{u - 1} \\
\end{array}
$$

Hence $I_r = u^{-r}.e^{u - 1}\big|_1^2 + r\int\limits_1^2u^{-r-1}.e^{u - 1} = \frac{e}{2^r} - 1 + rI_{r + 1}$

Therefore, since we'll be using this in a sum, we can notate the integral part as a sequence defined by the recursive formula:
$I_{r} = \frac{1}{r}(I_{r - 1} + 1 - \frac{e}{2^r})$

Now we can transfer that to the expression inside the sum and formally just find the sum of the new series we got:
$a_r = 1 - rI_r = 1 - (I_{r - 1} + 1 - \frac{e}{2^r}) = \frac{e}{2^r} - I_{r - 1}$ for $r > 0$.

So $a_{r + 1} = 1 - (r + 1)I_{r + 1} = \frac{e}{2^{r + 1}} - I_r = \frac{e}{2^{r + 1}} - \frac{1 - a_r}{r}$ for $r > 0$

So now we have:
$a_0 = 1$
$a_1 = \frac{e}{2} - I_0 = \frac{e}{2} - e + 1 = 1 - \frac{e}{2}$
$a_r = \frac{e}{2^r} - \frac{1 - a_{r-1}}{r - 1}$ for $r > 1$

And we want to find the sum of the aforementioned recursive sequence:
[tex]S = \sum_{r=0}^{\infty}a_r[/tex]

which I have really no idea how to achieve as of right now, but I guess the work til here narrows the problem down a bit. I'll have to go now, but will expect some comments. I suspect I've made a mistake somewhere since I didn't get something cancelling out in a sweet way

[peyo]

... I suspect I've made a mistake somewhere since I didn't get something cancelling out in a sweet way

I also think something earlier must be wrong, as the difference in values is visible here:
$a_1 = \frac{e}{2} - I_0 = \frac{e}{2} - e + 1 = 1 - \frac{e}{2}$

I calculate -0.125386 instead of -0.359141

After further check, this earlier formula is wrong:
$I_{r} = \frac{1}{r}(I_{r - 1} + 1 - \frac{e}{2^r})$

Using the following program, found that the correct recursive formula instead that match values is:

$I_{r} = \frac{1}{r-1}(I_{r - 1} + 1 - \frac{e}{2^(r-1)})$

Код: Избери целия код

from scipy.integrate import quad
import numpy as np

I_r = lambda x,r: 1 / (1 + math.log(x))**r

s = 0
I_prev = 0
I_rr = 0
for r in xrange(10):
    result = quad( I_r, 1, math.e, args = (r,))
    a_r = (1 - r*result[0])
   
   
    if r > 1:
        I_rr = (1/(r-1))*(I_prev +1 - math.e/(2**(r-1)))
    I_prev = result[0]
    s += a_r
    print ("r= %f I_r= %f I_rr= %f a_r= %f" %(r, result[0], I_rr, a_r))
print s
print ("1-math.e/2= %f" % ( 1-math.e/2,))

r= 0.000000 I_r= 1.718282 I_rr= 0.000000 a_r= 1.000000
r= 1.000000 I_r= 1.125386 I_rr= 0.000000 a_r= -0.125386
r= 2.000000 I_r= 0.766245 I_rr= 0.766245 a_r= -0.532490
r= 3.000000 I_r= 0.543337 I_rr= 0.543337 a_r= -0.630012
r= 4.000000 I_r= 0.401184 I_rr= 0.401184 a_r= -0.604736
r= 5.000000 I_r= 0.307823 I_rr= 0.307823 a_r= -0.539114
r= 6.000000 I_r= 0.244575 I_rr= 0.244575 a_r= -0.467452
r= 7.000000 I_r= 0.200350 I_rr= 0.200350 a_r= -0.402453
r= 8.000000 I_r= 0.168445 I_rr= 0.168445 a_r= -0.347559
r= 9.000000 I_r= 0.144728 I_rr= 0.144728 a_r= -0.302555
-2.95175678265
1-math.e/2= -0.359141



[peyo]

Also I can't directly use this program to get the final answer because of "RuntimeWarning: overflow encountered in double_scalars", but the answer is something less than -12.8814541698.

[Sup3rlum]

I think this integral is divergent. I've tried 3 different ways so far and it always tends towards negative infinity.


I'm a little bit late for the party, but this is what I've got:

I'll start off with our integral, but I'd use $n$ instead of $r$ if thats alright:

$I_n=\int_1^e\frac{1}{(1+lnx)^n}dx$

I'll use integration by parts to get a recursive expression:

$$
\begin{array}{ c| c| c }
&D&I \\
\hline
\\
+&(1+lnx)^{-n}&1 \\
\\
\hline
\\
-&\frac{-n(1+lnx)^{-n-1}}{x}&x \\
\\
\end{array}
$$

As an answer in recursive form we get:
$I_n=\bigg[x(1+lnx)^{-n}\bigg]_1^e+n\int_1^e(1+lnx)^{-(n+1)}dx$
$I_n=\frac{e}{2^n}-1+nI_{n+1}$

Rearrange:
$1-nI_{n+1}=\frac{e}{2^n}-I_n$
Subtract an $I_{n+1}$ term from both sides:

$1-nI_{n+1}-I_{n+1}=\frac{e}{2^n}-I_n-I_{n+1}$

Add up and adjust the index:

$1-(n+1)I_{n+1}=\frac{e}{2^n}-I_n-I_{n+1}$
$1-nI_n=\frac{e}{2^{n-1}}-I_{n-1}-I_n$

Now substitute into the summation:

$\sum_{n=0}^\infty(1-nI_n)=\sum_{n=0}^\infty\bigg(\frac{e}{2^{n-1}}-I_{n-1}-I_n\bigg)$

$\Rightarrow 4e-\sum_{n=0}^\infty(I_{n-1}+I_n)$

Now let's look at the right side of the difference, namely:

$\sum_{n=0}^\infty(I_{n-1}+I_n)=\sum_{n=0}^\infty\bigg(\int_1^e\frac{1}{(1+lnx)^{n-1}}dx+\int_1^e\frac{1}{(1+lnx)^{n}}dx\bigg)=\sum_{n=0}^\infty\bigg(\int_1^e\frac{2+lnx}{(1+lnx)^{n-1}}dx\bigg)$

The integrand is strictly positive in between these limits, so we can bring the summation inside, and use the infinite geometric sum with base $|(1+lnx)^{-1}| < 1$ for the bounds:

$=3e-1+\int_1^e\frac{2+lnx}{lnx}dx$

The geometric refactoring however is divergent. I might be wrong, but I haven't found any mistakes so far, and that is actually starting to seem reasonable.

[peyo]

I think this integral is divergent. I've tried 3 different ways so far and it always tends towards negative infinity.
...
The geometric refactoring however is divergent. I might be wrong, but I haven't found any mistakes so far, and that is actually starting to seem reasonable.

I've found a way to calculate the value of the sum with "r" up to 1e9 without loosing precision using the recursive integral and by the look of it, it seems divergent indeed.

figure_sumint1.png (22.91 KiB) Прегледано 475 пъти

Код: Избери целия код

from __future__ import division
import matplotlib.pyplot as plt
import math

I_prev = 1.1253860830832698
S_ar = 0.874613916917
r = 1
v2 = math.e/(2**(r-1))
X,Y = [],[]
for r in xrange(2,100000000):
    v2 /= 2
    I_r = (1/(r-1))*(I_prev + 1 - v2)
    a_r = (1 - r*I_r)
    S_ar += a_r
    I_prev = I_r
    #print ("r= %f I_r= %f a_r= %f S_ar= %f" %(r, I_r, a_r, S_ar))
    if r%1000000 == 0:
        X.append(r)
        Y.append(S_ar)
print S_ar

plt.plot(X,Y)

[Kre4etalo]

Guys, using your results, I think I've found a solution.
So, obviously, considering the integral form, $$\lim_{r\to\infty}I_r=0.$$
Using this limit and the recurrence formula $$I_r=\frac{1}{r-1}\left(I_{r-1}+1-\frac{e}{2^{r-1}}\right)$$
we obtain $$\lim_{r\to\infty}(r-1)I_r=\lim_{r\to\infty}I_{r-1}+1-\frac{e}{2^{r-1}}=1.$$
Thus $$\lim_{r\to\infty}rI_r=1.$$
Now observe that $$r(1-rI_r)=r\left(\frac{e}{2^{r-1}}-I_r-I_{r-1}\right),$$
hence $$\lim_{r\to\infty} r(1-rI_r)=\lim_{r\to\infty}\frac{e}{2^{r-1}}-\lim_{r\to\infty}rI_r-\lim_{r\to\infty}rI_{r-1}=0-1-1=-2.$$
This means that $$1-rI_r \sim \frac{-2}{r}$$
hence the original series are divergent.