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Test for convergence?
Summation 1 to infinity
3i/2^i
1 Answer
- 9 years agoFavorite Answer
Use the Ratio Test
a_i = 3i / 2^i
a_{i+1} = 3(i+1) / 2^(i+1)
Now find the ratio a_{i+1} / a_i
a_{i+1} / a_i = [ 3(i+1) / 2^(i+1) ] / [ 3i / 2^i ]
a_{i+1} / a_i = [ 3(i+1) 2^i ] / [ 2^(i+1) 3i ]
a_{i+1} / a_i = (i+1)/(2i)
Now take the limit, L, of the ratio as i→∞
L = lim_{i→∞} (i+1)/(2i) = 1/2
Since L=1/2 < 1, we have convergence.
As for what the series converges to, I cannot figure. Wolfram|Alpha says the value is 6, but I dont know how they figure this.