", B!TCHES!" C'mon, pra, close the deal.pra_ggresion wrote:sigma(1/2^n,1,infinity)<or = to integral(1/n^2,1,infinity)dn, let b=infinity
--> <or = to limit,b-->infinity of integral(1/n^2,1,b)dn
--> <or = to limit,b-->infinity of 1/n),1,b] by fundamental theorem of calculus
--> < or = to limit,b-->infinity of (-1/1)-(-1/b), 1 divided by infinity=0
Therefore sigma(1/2^n,1,infinity)<or= -1, assume equals b/c test for divergence sucks and it's a power series anyways.
e^(i*pi)=cos(pi)+i*sin(pi)=1
0.002+sigma(1/2^n,1,infinity)+e^(i*pi)=0.002-1+1
=$0.002
therefore, Randall got jipped .
pra_ggresion wrote:sigma(1/2^n,1,infinity)<or = to integral(1/n^2,1,infinity)dn, let b=infinity
--> <or = to limit,b-->infinity of integral(1/n^2,1,b)dn
--> <or = to limit,b-->infinity of 1/n),1,b] by fundamental theorem of calculus
--> < or = to limit,b-->infinity of (-1/1)-(-1/b), 1 divided by infinity=0
Therefore sigma(1/2^n,1,infinity)<or= -1, assume equals b/c test for divergence sucks and it's a power series anyways.
e^(i*pi)=cos(pi)+i*sin(pi)=1
0.002+sigma(1/2^n,1,infinity)+e^(i*pi)=0.002-1+1
=$0.002
therefore, Randall got jipped .
holddown wrote:Pra, you still have a sign error. Check Euler's identity.
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