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the sum of the series from 0 to infinity of (1+sin(n))/10^(n)
Do u have to find the exact value or just show that it converges?
It converges to zero I think.
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actually, i've got 10/9 + [10sin(1)]/[101-20cos(1)]
somebody really have to check if this is right, but here is the idea:
∑[1+sin(n)]/10^n=∑1/10^n+∑sin(n)/10^n the first term is clearly 10/9 (it's 1+0.1+0.01+ ....=1.1111....=10/9)
for the second term we write sin(n)=[e^(in) - e^(-in)]/2i so that the sum is [∑(e^(i)/10)^n - ∑(e^(-i)/10)^n]/2i which is 2 geometric series ... so after some calculation I've got [10sin(1)]/[101-20cos(1)]. Hence, the whole answer is 10/9 + [10sin(1)]/[101-20cos(1)].
Would help a lot if you put it in latex.
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lb sould be able to understand it, I'm sure
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