Question

double series

Answer 1

\[\huge \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}(-1)^{i+j}\frac{\ln(i+j)}{i+j}\]

Answer 2

i am looking for some symmetry,but i cannot see any

Answer 3

\[\sum_{i=1}^{\infty}((-1)^{1+i}\dfrac{\ln (1+i)}{1+i}+(-1)^{2+i}\dfrac{\ln (2+i)}{2+i}+...)\]
\[(\dfrac{\ln 2}{2}-\dfrac{\ln 3}{3}+\dfrac{\ln 4 }{4}-\dfrac{\ln 5}{5}+... )+(-\dfrac{\ln3}{3}+\dfrac{\ln 4}{4}-\dfrac{\ln (5)}{5}+\dfrac{\ln 6}{6}...)+...\]

Answer 4

https://www.facebook.com/groups/IMODYSM/

Answer 5

this is a high school olympiad problem

Answer 6

I was going to suggest splitting the series into odd/even indices, but I've never tried this with a double sum before...

Answer 7

something that seems significant to me is that the next series in line seems to lose the first term from the last:
\( \frac{\ln 2}{2} - \frac{\ln 3}{3} + \frac{\ln 4}{4} - \frac{\ln 5}{5} + \cdots \)
+ ( \( - \frac{\ln 3}{3} + \frac{\ln 4}{4} - \frac{\ln 5}{5} + \cdots \)
+ \( ( \frac{\ln 4}{4} - \frac{\ln 5}{5} \)
Does this seem useful? Or am I crazy and this is useless?

Answer 8

yes your right ,there seem to be a loss of terms,wich @SithsAndGiggles suggests we cud group by parity to make it look more readeable

Answer 9

even\[ \dfrac{\ln2 }{2}+5\dfrac{\ln4 }{4}+7\dfrac{\ln6 }{6}+...\]
odd \[ -(2\dfrac{\ln3 }{3}+4\dfrac{\ln5 }{5}+6\dfrac{\ln7 }{7}+...)\]

Answer 10

There was something wrong with what I previously wrote, as the sum I had very clearly diverged.

Answer 11

yes i think insteead of n its n-2

Answer 12

\[\huge \sum_{n=2}^{\infty} (n-1)(-1)^{n}\dfrac{\ln n}{n}\]

Answer 13

i think we are getting somwhere

\[\huge \sum_{n=1}^{\infty }(2n-1)\dfrac{\ln 2n}{2n}-\huge \sum_{n=1}^{\infty }(2n)\dfrac{\ln 2n+1}{2n+1}\]

Answer 14

The issue, is that both of those sums are divergent.

Answer 15

I'm pretty sure that the original series does not converge absolutely. Which means that it either diverges, or we aren't able to just start changing the order of summation however we want.

Answer 16

Is there any chance we could just compare this to the series
\[\sum_{n=1}^\infty (-1)^n\frac{\ln n}{n}~~?\]
I'm trying to get Wolfram to cooperate with a large number of terms for the partial sums, but it stops computing past a certain point.

Answer 17

So I don't really know if it converges (if it does, it must be conditional).

Answer 18

I'm relatively sure it converges conditionally by the alternating series test. At the very least, each of the sums over only \(j\) will converge.

As for any kind of comparison, I'm not seeing anything obvious. I've been trying to think about writing it as an integral, but that hasn't worked out very well either.

Answer 19

After simply sticking it into Mathematica, I get\[\sum _{m=1}^{\infty } \frac{1}{2} (-1)^m \left[-\gamma _1\left(\frac{m+1}{2}\right)+\gamma _1\left(\frac{m+2}{2}\right)+
\\+2 \log \left(\psi ^{(0)}\left(\frac{m+1}{2}\right)-\psi ^{(0)}\left(\frac{m}{2}+1\right)\right)\right]\]where the \(\gamma_1\) is the Stieltjes gamma and \(\psi^{(0)}\) is the polygamma function.

Unfortunately, I don't think this is helpful.

Answer 20

Something tells me those functions aren't in the curriculum for high school math...

Answer 21

If anything, it's harder to work with in that form.

Answer 22

Using Mathematica to do numerical approximations, I find that it's approximately equal to 0.065922448902297, but it also appears to converge exceptionally slowly.

Answer 23

yes the functions are too heavy to be high school content, wud suppose they where looking for an approaximation ,thank you @KingGeorge @SithsAndGiggles and @AccessDenied