If I’m understanding correctly, the argument here is:
A) limx→∞(∑∞n=1(ne−nxcos(−nx)))=−112
B) limx→∞(ne−nx)=n
C) limx→∞(cos(−nx))=1
Therefore, ∑∞n=1(n∗1)=−112.
First off, this seems to have an implicit assumption that limx→∞(∑∞n=1(f(x,n)∗g(x,n)))=∑∞n=1((limx→∞f(x,n))∗(limx→∞g(x,n))).
I think this assumption is true for any functions f and g, but I’ve learned not to always trust my intuitions when it comes to limits and infinity; can anyone else confirm this is true?
Second, A seems to depend on the relative sizes of the infinities, so to speak. If j and k are large but finite numbers, then ∑jn=1(ne−nkcos(−nk))≈−112 if and only if j is substantially greater than k; if k is close to or larger than j, it becomes much less than or greater than −1/12.
I’m not sure exactly how this works when it comes to infinities—does the infinity on the sum have to be larger than the infinity on the limit for this to hold? I’m pretty sure what I just said was nonsense; is there a non-nonsensical version?
In conclusion, I don’t know how infinities work and hope someone else does.
I didn’t make any claim about limits. If you’re looking for rigor, you’re in the wrong place, as I tried to make clear in the introduction.
But (A) is true without any unconventional weirdness: ∑∞k=1kekxcos(kx)=e(1+i)x(e2ix−4e(1+i)x+e2x+e(2+2i)x+1)2(−ex+eix)2(−1+e(1+i)x)2 (from Mathematica), and limx→0− of that is −112.
If I’m understanding correctly, the argument here is:
A) limx→∞(∑∞n=1(ne−nxcos(−nx)))=−112
B) limx→∞(ne−nx)=n
C) limx→∞(cos(−nx))=1
Therefore, ∑∞n=1(n∗1)=−112.
First off, this seems to have an implicit assumption that limx→∞(∑∞n=1(f(x,n)∗g(x,n)))=∑∞n=1((limx→∞f(x,n))∗(limx→∞g(x,n))).
I think this assumption is true for any functions f and g, but I’ve learned not to always trust my intuitions when it comes to limits and infinity; can anyone else confirm this is true?
Second, A seems to depend on the relative sizes of the infinities, so to speak. If j and k are large but finite numbers, then ∑jn=1(ne−nkcos(−nk))≈−112 if and only if j is substantially greater than k; if k is close to or larger than j, it becomes much less than or greater than −1/12.
I’m not sure exactly how this works when it comes to infinities—does the infinity on the sum have to be larger than the infinity on the limit for this to hold? I’m pretty sure what I just said was nonsense; is there a non-nonsensical version?
In conclusion, I don’t know how infinities work and hope someone else does.
I didn’t make any claim about limits. If you’re looking for rigor, you’re in the wrong place, as I tried to make clear in the introduction.
But (A) is true without any unconventional weirdness: ∑∞k=1kekxcos(kx)=e(1+i)x(e2ix−4e(1+i)x+e2x+e(2+2i)x+1)2(−ex+eix)2(−1+e(1+i)x)2 (from Mathematica), and limx→0− of that is −112.