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

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

    \[ \sum_{n=1}^{\infty} \int_n^{n+1} e^{-\sqrt{x}} \, dx. \]


First, we make the substitution s = \sqrt{x}, which gives us ds = \frac{1}{2 \sqrt{x}} \, dx. Then we have

    \[ \int e^{-\sqrt{x}} \, dx = 2 \int se^{-s} \, ds. \]

This integral we evaluate using integration by parts with

    \begin{align*}  u &= s & du &= ds \\  dv &= e^{-s} \, ds & v &= -e^{-s}.  \end{align*}

Therefore, we have

    \begin{align*}  \int e^{-\sqrt{x}} \, dx &= 2 \int se^{-s} \, ds \\[9pt]  &= 2 \left( -se^{-s} + \int e^{-s} \, ds \right) \\[9pt]   &= 2 (e^{-s}(-1-s)) \\  &= -2e^{-\sqrt{x}} (1+\sqrt{x}). \end{align*}

So, for the definite integral from n to n+1 we have

    \begin{align*}   \int_n^{n+1} e^{-\sqrt{x}} \, dx &= -2e^{-\sqrt{n+1}} (1+\sqrt{n+1}) + 2 e^{-\sqrt{n}}(1+\sqrt{n}) \\[9pt]  &= 2e^{-\sqrt{n}} (1+\sqrt{n}) - 2e^{-\sqrt{n+1}} (1+\sqrt{n+1}). \end{align*}

But then the series is a telescoping series with

    \[ b_n = 2e^{-\sqrt{n}} (1+\sqrt{n}) \quad \implies \quad \lim_{n \to \infty} b_n = 0. \]

Hence,

    \[ \sum_{n=1}^{\infty} (b_n - b_{n+1}) = b_1 - L = \frac{4}{e}. \]

Hence, the series converges.

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