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Conclude if the given series converges or diverges and justify the conclusion

by RoRi

Test the following series for convergence or divergence. Justify the decision.

    \[ \sum_{n=1}^{\infty} \frac{n^2}{2^n}. \]

The series converges. To show this we apply the integral test (Theorem 10.11 on page 397 of Apostol), letting

    \[ f(x) = \frac{x^2}{2^x}, \qquad t_n = \int_1^n \frac{x^2}{2^x} \, dx. \]

As in the previous exercise we evaluate the integral using integration by parts with

    \begin{align*} u &= x^2, & du &= 2x \, dx \\ dv &= 2^{-x} \, dx & v &= \frac{-1}{2^x \log 2}. \end{align*}

Therefore,

    \begin{align*} \int_1^n \frac{x^2}{2^x} \, dx &= \frac{-x^2}{2^x \log 2} \Bigr \rvert_1^n + \frac{2}{\log 2} \int_1^n \frac{x}{2^x} \, dx \\[9pt] &= \frac{-n^2}{2^n \log 2} + \frac{1}{2 \log 2} + \frac{2}{\log 2} \int_1^n \frac{x}{2^x} \, dx. \end{align*}

But, from the previous exercise we know

    \[ \lim_{n \to \infty} \int_1^n \frac{x^}{2^x} \, dx \]

converges; hence,

    \[ \sum_{n=1}^{\infty} \frac{n^2}{2^n} \]

convergges.

2 comments

  1. Anonymous says:

    Very nice proof!

    I offer an alternative:

    Since \lim_{n \to \infty} \frac{\frac{n^2}{2^n}}{\frac{1}{n^2}} = 0 the series \sum_{n=1}^{\infty} \frac{n^2}{2^n} converges from T10.9 (below is the case for the limit being 0) since \sum_{n=1}^{\infty} \frac{1}{n^2} is the Riemann Zeta function \zeta(s) which converges.

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