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anonymous
 one year ago
Identify whether the series summation of 15 open parentheses 4 close parentheses to the i minus 1 power from 1 to infinity is a convergent or divergent geometric series and find the sum, if possible.
anonymous
 one year ago
Identify whether the series summation of 15 open parentheses 4 close parentheses to the i minus 1 power from 1 to infinity is a convergent or divergent geometric series and find the sum, if possible.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{\infty}^{i=1}15(4)^i1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's suppose to be 15(4)^i=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a convergent geometric series. The sum is –5. This is a divergent geometric series. The sum is –5. This is a convergent geometric series. The sum cannot be found. This is a divergent geometric series. The sum cannot be found.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0the exponent is `i1` ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whether you meant \(5(4)^{i+1}\) or \(5(4)^{i1}\), consider what happens as \(i\to\infty\). \[\lim_{i\to\infty}5(4)^{i}=\infty\neq0\] What do you know about series of the form \(\sum a_n\) for which \(\lim\limits_{n\to\infty}a_n\neq0\)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes the exponent is i1, and I think it's divergent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I couldnt find the sum so D!

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0Yep whenever it's divergent, the infinite number of terms don't sum to one fixed number.
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