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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.

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

The series converges. We apply the integral test (Theorem 10.11 on page 397 of Apostol), letting

    \[ f(x) = \frac{x+1}{2^x}, \qquad t_n = \int_1^n \frac{x+1}{2^x} \, dx. \]

To evaluate the integral we use integration by parts with

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

Then we have

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

Therefore, by the integral test we have

    \begin{align*} \lim_{n \to \infty} \{ t_n \} &= \lim_{n \to \infty} \left( \frac{-(n+1)}{2^n \log 2} + \frac{1}{\log 2} + \frac{1}{2 (\log 2)^2} - \frac{n}{2^n (\log 2)^2} \right) \\[9pt] &= \frac{1}{\log 2} + \frac{1}{2(\log 2)^2}. \end{align*}

Hence, the sequence \{ t_n \} converges; therefore, \{ s_n \} converges which implies the convergence of

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

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