Use the Integral Test to determine the convergence or divergence of the series. $$\sum _ { n = 1 } ^ { \infty } n e ^ { - \frac { n } { 2 } }$$
A) converges
B) diverges
C) Integral Test inconclusive

Question

Quizplus · Accepted Answer

Using the Integral Test, we have:

$\int_{1}^{\infty}ne^{-\frac{n}{2}}\mathrm{d}n = \left[-2ne^{-\frac{n}{2}}ight]_{1}^{\infty} + 2\int_{1}^{\infty}e^{-\frac{n}{2}}\mathrm{d}n$

Since $\lim_{n 	o \infty} n e^{-\frac{n}{2}} = 0$ (using L'Hopital's rule or algebraic manipulation), the first term evaluates to 2. The second integral evaluates to $2\left[-2e^{-\frac{n}{2}}ight]_{1}^{\infty} = 4e^{-\frac{1}{2}}$, which is finite.

Thus, since the integral $\int_{1}^{\infty}ne^{-\frac{n}{2}}\mathrm{d}n$ converges, by the Integral Test, the series $\sum_{n=1}^{\infty}ne^{-\frac{n}{2}}$ also converges. Therefore, the answer is A: converges.

Use the Integral Test to Determine the Convergence or Divergence ∑n=1∞ne−n2\sum _ { n = 1 } ^ { \infty } n e ^ { - \frac { n } { 2 } }n=1∑∞​ne−2n​

Use the Integral Test to Determine the Convergence or Divergence $\sum _ { n = 1 } ^ { \infty } n e ^ { - \frac { n } { 2 } }$