Use the Integral test to determine whether the series is convergent or divergent.

Question

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

Answer 1

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.