Asked by Philip Martinson
Use the Integral test to determine whether the series is convergent or divergent.
infinity "series symbol" n=1 (ne^(n"pi"))
Note:
I don't know how to solve or work out so show all your work. And give the answer in EXACT FORM example 3pi, sqrt(2), ln(2) not decimal approximations like 9.424,1.4242,1232
infinity "series symbol" n=1 (ne^(n"pi"))
Note:
I don't know how to solve or work out so show all your work. And give the answer in EXACT FORM example 3pi, sqrt(2), ln(2) not decimal approximations like 9.424,1.4242,1232
Answers
Answered by
oobleck
The integral test says that
∑ne^(nπi) converges if and only if ∫ xe^(xπi) dx converges
That integral diverges, since
e^(xπi) = cos(πx) + i sin(πx)
Since those oscillate, between -1 and 1, the integral does not converge.
∑ne^(nπi) converges if and only if ∫ xe^(xπi) dx converges
That integral diverges, since
e^(xπi) = cos(πx) + i sin(πx)
Since those oscillate, between -1 and 1, the integral does not converge.
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