Calc6.4

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Discuss the convergence of \sum _{{n=1}}^{{\infty }}{\frac  {n}{e^{n}}}.

We can use the integral test on this one because the function is continuous, positive, and decreasing. To do the integral, we will use integration by parts.

\int _{{1}}^{{\infty }}{\frac  {x}{e^{x}}}\,dx\,

u=x\,

du=dx\,

dv=e^{{-x}}\,dx\,

v=-e^{{-x}}\,

\int _{{1}}^{{\infty }}{\frac  {x}{e^{x}}}\,dx=-xe^{{-x}}+\int _{{1}}^{{\infty }}e^{{-x}}\,dx=\left[-xe^{{-x}}-e^{{-x}}\right]_{{1}}^{{\infty }}\,

Taking the limit as x\rightarrow \infty and plugging in 1 gives

\int _{{1}}^{{\infty }}{\frac  {x}{e^{x}}}\,dx={\frac  {-2}{e}}\,

Thus, by the integral test, since this converges, the series also converges.


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