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Math Help - Improper Integral, Sequences, and Series

  1. #1
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    Improper Integral, Sequences, and Series

    Hey guys, I have a few problems that if possible, I would love to have someone check my solutions to them.

    Problem 1. Determine whether the improper integral \int^{\infty}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx converges, or diverges. In the case of convergence, give its value.

    My Solution


    = \lim_{t \to \infty}\int^{t}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx = \lim_{t \to \infty}[lnx - ln(x^2 + 1)]^{t}_{1} = \lim_{t \to \infty}[(lnt - ln(t^2 + 1) - (ln1 - ln2)] = \lim_{t \to \infty}(lnt - ln(t^2 + 1) + ln2) = \infty - \infty + ln2 = ln2 ... Converges to ln2

    Problem 2.
    Determine whether the sequence with the given general term converges, or diverges. In the case of convergence, give its value.

    a.) a_{n} = \frac{2n - 1}{3 - 5n}

    My Solution
    <br />
\lim_{n \to \infty}\frac{2n - 1}{3 - 5n} = \frac{\infty}{\infty}, Using L'Hopital's Rule. = \lim_{n \to \infty}\frac{-2}{5} = \frac{-2}{5} ... Converges to \frac{-2}{5}

    b.) a_{n} = (-1)^{n}\sin{\frac{n\pi}{2}}

    My Solution

    \lim_{n \to \infty}|(-1)^{n}\sin{\frac{n\pi}{2}}| = \lim_{n \to \infty}|(-1)^{n}|*|\sin{\frac{n\pi}{2}}| = \lim_{n \to \infty}1 * \sin{\frac{n\pi}{2}} = No Limit, Diverges

    c.) a_{n} = (1 + \frac{1}{n})^n

    My Solution

    1^{\infty} ... y = (1 + \frac{1}{n})^{n}, lny = nln(1 + \frac{1}{n})

    \lim_{n \to \infty} lny = \lim_{n \to \infty}\frac{ln(1 + \frac{1}{n})}{\frac{1}{n}} = \frac{0}{0}, Using L'Hopital's Rule.  = \lim_{n \to \infty}\frac{\frac{-1}{n^2 + n}}{\frac{-1}{n^2}} = \lim_{n \to \infty}\frac{n}{n + 1} = \frac{\infty}{\infty} = \lim_{n \to \infty}\frac{1}{1} = 1 ... lny \to 1, y \to e ... Converges to e

    Problem 3. Calculate the sum of the given convergent geometric series.

    a.) \frac{9}{4} - \frac{3}{2} + 1 ...

    My Solution

    a = \frac{9}{4}, r = \frac{-2}{3}

    \frac{\frac{9}{4}}{1 + \frac{2}{3}} = \frac{27}{20}

    b.)
    \Sigma^{\infty}_{n = 1}(-1)^{n}(\frac{3}{7})^n

    My Solution

    a = \frac{-3}{7}, r = \frac{-3}{7}

    \frac{\frac{-3}{7}}{1 + \frac{3}{7}} = \frac{-3}{10}

    Problem 4. Write the first three terms of the sequence of partial sums of the series \Sigma^{\infty}_{k = 1}\frac{k}{2k - 1}.

    My Solution

    S_{1} = \frac{1}{1} = 1
    S_{2} = 1 + \frac{2}{3} = \frac{5}{3}
    S_{3} = \frac{5}{3} + \frac{3}{5} = \frac{34}{15}

    1, \frac{5}{3}, \frac{34}{15}

    Thank you in advance for any help!
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  2. #2
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    Quote Originally Posted by mturner07 View Post
    Hey guys, I have a few problems that if possible, I would love to have someone check my solutions to them.

    Problem 1. Determine whether the improper integral \int^{\infty}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx converges, or diverges. In the case of convergence, give its value.

    My Solution

    = \lim_{t \to \infty}\int^{t}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx = \lim_{t \to \infty}[lnx - ln(x^2 + 1)]^{t}_{1} = \lim_{t \to \infty}[(lnt - ln(t^2 + 1) - (ln1 - ln2)] = \lim_{t \to \infty}(lnt - ln(t^2 + 1) + ln2) = \infty - \infty + ln2 = ln2 ... Converges to ln2


    Nop: \ln t-\ln(t^2+1)=\ln\left(\frac{t}{t^2+1}\right)\xrighta  rrow[t\to\infty]{}-\infty and thus the integral diverges.

    Of course, you can't write \infty-\infty=0 ...

    Problem 2. Determine whether the sequence with the given general term converges, or diverges. In the case of convergence, give its value.

    a.) a_{n} = \frac{2n - 1}{3 - 5n}

    My Solution
    <br />
\lim_{n \to \infty}\frac{2n - 1}{3 - 5n} = \frac{\infty}{\infty}, Using L'Hopital's Rule. = \lim_{n \to \infty}\frac{-2}{5} = \frac{-2}{5} ... Converges to \frac{-2}{5}


    Nop. You can't use DIRECTLY L'Hospital with a discrete variable since L'H implies the use of derivative which use limits which need a continuous variable. Of course, you can use L'H with \frac{2x-1}{3-5x} and then use that the limit stays the same no matter how x\rightarrow \infty , and thus this is so if you choose to go to the limit along the naturals.

    Another way, perhaps more natural , to divide both numerator and denominator by the highest power of n and use arithmetic of limits:

    \frac{2n-1}{3-5n}\cdot\frac{1\slash n}{1\slash n} = \frac{2-\frac{1}{n}}{\frac{3}{n}-5}\xrightarrow [n\to\infty]{}\frac{2-0}{0-5}=-\frac{2}{5}

    Tonio


    b.) a_{n} = (-1)^{n}\sin{\frac{n\pi}{2}}

    My Solution

    \lim_{n \to \infty}|(-1)^{n}\sin{\frac{n\pi}{2}}| = \lim_{n \to \infty}|(-1)^{n}|*|\sin{\frac{n\pi}{2}}| = \lim_{n \to \infty}1 * \sin{\frac{n\pi}{2}} = No Limit, Diverges

    c.) a_{n} = (1 + \frac{1}{n})^n

    My Solution

    1^{\infty} ... y = (1 + \frac{1}{n})^{n}, lny = nln(1 + \frac{1}{n})

    \lim_{n \to \infty} lny = \lim_{n \to \infty}\frac{ln(1 + \frac{1}{n})}{\frac{1}{n}} = \frac{0}{0}, Using L'Hopital's Rule.  = \lim_{n \to \infty}\frac{\frac{-1}{n^2 + n}}{\frac{-1}{n^2}} = \lim_{n \to \infty}\frac{n}{n + 1} = \frac{\infty}{\infty} = \lim_{n \to \infty}\frac{1}{1} = 1 ... lny \to 1, y \to e ... Converges to e

    Problem 3. Calculate the sum of the given convergent geometric series.

    a.) \frac{9}{4} - \frac{3}{2} + 1 ...

    My Solution

    a = \frac{9}{4}, r = \frac{-2}{3}

    \frac{\frac{9}{4}}{1 + \frac{2}{3}} = \frac{27}{20}

    b.) \Sigma^{\infty}_{n = 1}(-1)^{n}(\frac{3}{7})^n

    My Solution

    a = \frac{-3}{7}, r = \frac{-3}{7}

    \frac{\frac{-3}{7}}{1 + \frac{3}{7}} = \frac{-3}{10}

    Problem 4. Write the first three terms of the sequence of partial sums of the series \Sigma^{\infty}_{k = 1}\frac{k}{2k - 1}.

    My Solution

    S_{1} = \frac{1}{1} = 1
    S_{2} = 1 + \frac{2}{3} = \frac{5}{3}
    S_{3} = \frac{5}{3} + \frac{3}{5} = \frac{34}{15}

    1, \frac{5}{3}, \frac{34}{15}

    Thank you in advance for any help!
    .
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