For all ∣x∣<1
n=1∑∞m=1∑∞mxn(mn+m)−1=log(1−x1)
Convergence Test
Ratio Test
We check if the following limit exists and results in a finite value
n→∞limm→∞liman−1,m−1an,m
where
an,m=m(mn+m)xn=(n+m)!n! (m−1)!xn
∴n→∞limm→∞lim((n+m)!n! (m−1)!xn)((n−1)! (m−2)!(n+m−2)!xn−11)===n→∞limm→∞lim(n+m)(n+m−1)xn(m−1)\nonumbern→∞limm→∞limmnx−m22n2x+m3n2(3n−1)x+O(m41)\nonumber0\nonumber
Thus, we can conclude via the ratio test that the series converges.
We observe the denominator closely.
m(mn+m)1=(n+m)!n! (m−1)!=B(m,n+1)=∫01tn(1−t)m−1 dt
where B(x,y) denotes the Beta function.
Thus, we can express the sum as
n=1∑∞m=1∑∞m(mn+m)xn==n=1∑∞m=1∑∞∫01(xt)n(1−t)m−1 dt\nonumber∫01(n=1∑∞(xt)n)(m=1∑∞(1−t)m−1) dt\nonumber
Now, since x ∈(−1,1), we can use the Geometric Expansion that is ∑n=1∞rn=1−rr, to condense the sum as
∫01(n=1∑∞(xt)n)(m=1∑∞(1−t)m−1) dt==∫01(1−xtxt)(t1) dt\nonumber∫011−xtx dt\nonumber
The only work left is to evaluate the integral
∫011−xtx dt==−log(1−xt)∣01\nonumberlog(1−x1)\nonumber
And thus,
n=1∑∞m=1∑∞m(mn+m)xn=log(1−x1)
Proof By Jon Haussmann
n=1∑∞m=1∑∞m(mn+m)xn==n=1∑∞xnm=1∑∞(m+n)!(m−1)! n!\nonumbern=1∑∞nxnm=1∑∞(m+n)!(m−1)! n! n\nonumber
Now, we try telescoping the following series :
m=1∑∞(m+n)!(m−1)! n! n∴n=1∑∞m=1∑∞m(mn+m)xn=======m=1∑∞(m+n)!(m−1)! n! ((m+n)−m)\nonumberm=1∑∞(m+n)!(m−1)! n! (m+n)−m=1∑∞(m+n)!(m−1)! n! m\nonumberm=1∑∞(m+n−1)!(m−1)! n!−m=1∑∞(m+n)!m! n!\nonumbern!(m=1∑∞(m+n−1)!(m−1)!−m=1∑∞(m+n)!m!)\nonumbern!(n!0!)\nonumber1\nonumbern=1∑∞nxn\nonumber
And thus,
n=1∑∞m=1∑∞m(mn+m)xn=log(1−x1)